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8.1 CONGRUENT AND SIMILAR FIGURES:

Two geometric figures having the same shape and size are known as congruent figures. Geometric figures having the same shape but different sizes are known as similar figures.

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8.2 SIMILAR TRIANGLES:

Two triangles ABC and DEF are said to be similar if their

(i) Corresponding angles are equal. i.e. $\mathrm{\angle }A=\mathrm{\angle }D,\mathrm{\angle }B=\mathrm{\angle }E,\mathrm{\angle }C=\mathrm{\angle }F$ And,

(ii) Corresponding sides are proportional i.e. $\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$

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8.2 (a) Characteristic Properties of Similar Triangles :

(i) (AAA Similarity) If two triangles are equiangular, then they are similar.

(ii) (SSS Similarity) If the corresponding sides of two triangles are proportional, then they are similar.

(iii) (SAS Similarity) If in two triangle’s one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar.

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8.2 (b) Results Based Upon Characteristic Properties of Similar Triangles :

(i) If two triangles are equiangular, then the ratio of the corresponding sides is the same as the ratio of the corresponding medians.

(ii) If two triangles are equiangular, then the ratio of the corresponding sides is same at the ratio of the corresponding angle bisector segments.

(iii) if two triangles are equiangular then the ratio of the corresponding sides is same at the ratio of the corresponding altitudes.

(vi) If one angle of a triangle is equal to one angle of another triangle and the bisectors of these equal angles divide the opposite side in the same ratio, then the triangles are similar.

(v) If two sides and a median bisecting one of these sides of a triangle are respectively proportional to the two sides and the corresponding median of another triangle, then the triangles are similar.

(vi) If two sides and a median bisecting the third side of a triangle are respectively proportional to the corresponding sides and the median another triangle, then two triangles are similar.

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8.3 THALES THEOREM (BASIC PROPROTIONALITY THEOREM) :

Statement: If a line is drawn parallel to one side of a triangle to intersect the other sides in distinct points, then the other two sides are divided in the same ratio.

Given: $$ A triangle $ABC$ in which a line parallel to side $BC$ intersects other two sides $AB$ and $AC$ at $D$ and E respectively.

To Prove :

$\frac{AD}{DB}=\frac{AE}{EC}$

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Construction : Join $BE$ and $CD$ and draw $DM\perp AC$ and $EN\perp AB$.

Proof : $$ Area of $\mathrm{△}ADE(=\frac{1}{2}.$ base $\times$ height $)=\frac{1}{2}AD\times EN$.

Area of $\mathrm{△}ADE$ is denoted as are (ADE)

So, $ar(ADE)=\frac{1}{2}DB\times EN$ And $ar(BDE)=\frac{1}{2}DB\times EN$,

Therefore, $\frac{ar(ADE)}{ar(BDE)}=\frac{\frac{1}{2}AD\times EN}{\frac{1}{2}DB\times EN}=\frac{AD}{DB}$ …(i)

Similarly, $ar(ADE=\frac{1}{2}AE\times DM.$ and $ar(DEC=\frac{1}{2}EC\times DM.$.

And

$\frac{ar(ADE)}{ar(DEC)}=\frac{\frac{1}{2}AE\times DM}{\frac{1}{2}EC\times DM}=\frac{AE}{EC}$ …(ii)

Note that $\mathrm{△}BDE$ and $\mathrm{△}DEC$ are on the same base DE and between the two parallel lines $BC$ and $DE$.

So, $ar(BDE)=ar(DEC)$ …(iii)

Therefore, from (i), (ii) and (iii), we have :

$\frac{AD}{DB}=\frac{AE}{EC}\text{ Hence Proved. }$

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Corollary : If in a $\mathrm{△}ABC$, a line $DE||BC$, intersects $AB$ in $D$ and $AC$ in $E$, then

(i) $\frac{DB}{AD}=\frac{EC}{AE}$

(ii) $\frac{AB}{AD}=\frac{AC}{AE}$

(ii) $\frac{AD}{AB}=\frac{AE}{AC}$

(iv) $\frac{AB}{DB}=\frac{AC}{EC}$

(v) $\frac{DB}{AB}=\frac{EC}{AC}$

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8.3 (a) Converse of Basic Proportionality Theorem :

If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

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8.3 (b) Some Important Results and Theorems :

(i) The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

(ii) In a triangle $\mathbf{A}\mathbf{B}\mathbf{C}$, if $\mathbf{D}$ is a point on $\mathbf{B}\mathbf{C}$ such that $\mathbf{D}$ divides $\mathbf{B}\mathbf{C}$ in the ratio $\mathbf{A}\mathbf{B}:\mathbf{A}\mathbf{C}$, then $\mathbf{A}\mathbf{D}$ is the bisector of $\mathrm{\angle }A$.(iii) The external bisector of an angle of a triangle divides the opposite sides externally in the ratio of the sides containing the angle.

(iv) The line drawn from the mid-point of one side of a triangle parallel to another side bisects the third side. (v) The line joining the mid-points of two sides of a triangle is parallel to the third side.

(vi) The diagonals of a trapezium divide each other proportionally.

(vii) If the diagonals of a quadrilateral divide each other proportionally, then it is a trapezium.

(viii) Any line parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.

(ix) If three or more parallel lines are intersected by two transversal, then the intercepts made by them on the transversal are proportional.

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Ex. 1 In a $\mathrm{△}ABC,D$ and $E$ are points on the sides $AB$ and $AC$ respectively such that $DE|BC$. If $AD=4x-3$,

$AE=8x-7,BD=3x-1$ and $CE=5x-3$, find the value of $x$. $$ [CBSE - 2006]

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Sol. In $\mathrm{△}ABC$, we have

$DE||BC$

$\begin{array}{cc}\therefore & \frac{AD}{DB}=\frac{AE}{EC}\text{ [By Basic Proportionality Theorem] }\\ \Rightarrow & \frac{4x-3}{3x-1}=\frac{8x-7}{5x-3}\\ \Rightarrow & 20x^2-15x-12x+9=24x^2-21x-8x+7\\ \Rightarrow & 20x^2-27x+9=24x^2-29x+7\\ \Rightarrow & 4x^2-2x-2=0\\ \Rightarrow & 2x^2-x-1=0\\ \Rightarrow & (2x+1)(x-1)=0\\ \Rightarrow & x=1\text{ or }x=-\frac{1}{2}\end{array}$

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So, the required value of $x$ is 1 .

$[x=-\frac{1}{2}\text{ is neglected as length can not be negative }].$

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Ex. 2 $D$ and $E$ are respectively the points on the sides $AB$ and $AC$ of a $\mathrm{△}ABC$ such that $AB=12cm,AD=8cm$, $AE=12cm$ and $AC=18cm$, show that $DE||BC$.

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Sol. We have,

$AB=12cm,AC=18m,AD=8cm$ and $AE=12cm$.

$\therefore BD=AB-AD=(12-8)cm=4cm$

$CE=AC-AE=(\begin{array}{cc}18& 12\end{array})cm=6cm$

Now, $\frac{AD}{BC}=\frac{8}{4}=\frac{2}{1}$

And, $\frac{AE}{CE}=\frac{12}{6}=\frac{2}{1}\Rightarrow \frac{AD}{BD}=\frac{AE}{CE}$

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Thus, $DE$ divides sides $AB$ and $AC$ of $\mathrm{△}ABC$ in the same ratio. Therefore, by the conserve of basic proportionality theorem we have $DE||BC$.

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Ex. 3 In a trapezium $ABCDAB||DC$ and $DC=2AB$. EF drawn parallel to $AB$ cuts $AD$ in $F$ and $BC$ in $E$ such that $\frac{BE}{EC}=\frac{3}{4}$. Diagonal $DB$ intersects $EF$ at $G$. Prove that $7FE=10AB$.

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Sol.

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From (i) and (ii), we get

$\frac{FG}{AB}=\frac{4}{7}$ i.e. $FG=\frac{4}{7}AB$ …(iii)

In $\mathrm{△}BEG$ and $\mathrm{△}BCD$, we have

$\mathrm{\angle }BEG=\mathrm{\angle }BCD$ [Corresponding angle $\therefore EG||CD$ ]

$\mathrm{\angle }GBE=\mathrm{\angle }DBC$ [Common]

$\therefore \mathrm{△}BEG\sim \mathrm{△}BCD$ [By AA rule of similarity]

$\therefore \frac{BE}{BC}=\frac{EG}{CD}$

$\therefore \frac{3}{7}=\frac{EG}{CD}[\because \frac{BE}{EG}=\frac{3}{7}.$ i.e.. $.\frac{EC}{BE}=\frac{4}{3}\Rightarrow \frac{EC+BE}{BE}=\frac{4+3}{3}]\Rightarrow \frac{BC}{BE}=\frac{7}{3}$

$\therefore EG=\frac{3}{7}CD=\frac{3}{7}(2AB)[\because CD=2AB$ (given) $]$

$\therefore EG=\frac{6}{7}AB$ …(iv)

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Adding (iii) and (iv), we get

$FG+EG=\frac{4}{7}AB+\frac{6}{7}AB=\frac{10}{7}AB$

$\Rightarrow EF=\frac{10}{7}AB$ i.e., $7EF=10AB$.

Hence proved.

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Ex. 4 In $\mathrm{△}ABC$, if $AD$ is the bisector of $\mathrm{\angle }A$, prove that $\frac{\text{ Area }(\mathrm{△}ABD)}{\text{ Area }(\mathrm{△}ACD)}=\frac{AB}{AC}$

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Sol. In $\mathrm{△}ABC,AD$ is the bisector of $\mathrm{\angle }A$.

$\therefore \frac{AB}{AC}=\frac{BD}{DC}$ ….(i) [By internal bisector theorem]

From A draw $AL\perp BC$

$\therefore \frac{\text{ Area }(\mathrm{△}ABD)}{\text{ Area }(\mathrm{△}ACD)}=\frac{\frac{1}{2}BD\cdot AL}{\frac{1}{2}DC\cdot AL}=\frac{BD}{DC}=\frac{AB}{AC}$

[From (i)]

Hence Proved.

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Ex. 5 $\mathrm{\angle }BAC={90}^0,AD$ is its bisector. IF $DE\perp AC$, prove that $DE\times (AB+AB)=AB\times AC$.

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Sol. It is given that $AD$ is the bisector of $\mathrm{\angle }A$ of $\mathrm{△}ABC$.

$\therefore \frac{AB}{AC}=\frac{BD}{DC}$

$\Rightarrow \frac{AB}{AC}+1=\frac{BD}{DC}+1\text{ [Adding }1\text{ on both sides] }$

$\Rightarrow \frac{AB+AC}{AC}=\frac{BD+DC}{DC}$

$\Rightarrow \frac{AB+AC}{AC}=\frac{BC}{DC}$ …(i)

In ${\mathrm{\Delta }}^{\prime }$ s CDE and CBA, we have

$\mathrm{\angle }DCE=\mathrm{\angle }BCA$ [Common]

$\mathrm{\angle }DEC=\mathrm{\angle }BAC$ [Each equal to ${90}^0$ ]

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So, by AA-criterion of similarity

$\mathrm{\Delta }CDE\sim \mathrm{\Delta }CBA$

$\Rightarrow \frac{CD}{CB}=\frac{DE}{BA}\Rightarrow \frac{AB}{DE}=\frac{BC}{DC}$ …(ii)

From (i) and (ii), we have

$\Rightarrow \frac{AB+AC}{AC}=\frac{AB}{DE}\Rightarrow DE\times (AB+AC)=AB\times AC$

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Ex. 6 In the given figure, $PA,QB$ and $RC$ are each perpendicular to $AC$. Prove that $\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$

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Sol. In $\mathrm{△}PAC$, we have $BQ||AP$

$\begin{array}{rl}& \Rightarrow \frac{BQ}{AP}=\frac{CB}{CA}\\ & [\therefore \mathrm{\Delta }CBQ\sim \mathrm{\Delta }CAP]\\ & \Rightarrow \frac{y}{x}=\frac{CB}{CA}\dots (i)\\ & \text{ In }\mathrm{△}ACR\text{, we have }BQ||CR\\ & \Rightarrow \frac{BQ}{CR}=\frac{AB}{AC}[\therefore \mathrm{\Delta }ABQ\sim \mathrm{\Delta }ACR]\\ & \Rightarrow \frac{y}{z}=\frac{AB}{AC}\dots (ii)\end{array}$

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$\begin{array}{rl}& \text{ Adding (i) and (ii), we get }\\ & \frac{y}{x}+\frac{y}{z}=\frac{CB}{AC}+\frac{AB}{AC}\\ & \Rightarrow \frac{y}{x}+\frac{y}{z}=\frac{AB+BC}{AC}& \Rightarrow \frac{y}{x}+\frac{y}{z}=\frac{AC}{AC}\\ & \Rightarrow \frac{y}{x}+\frac{y}{z}=1& \Rightarrow \frac{1}{x}+\frac{1}{z}=\frac{1}{y}\end{array}$

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Ex. 7 In the given figure, $AB||CD$. Find the value of $x$.

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Sol. Since the diagonals of a trapezium divide each other proportionally.

$\begin{array}{cc}\therefore & \frac{AO}{OC}=\frac{BO}{OD}\\ \Rightarrow & \frac{3x-19}{x-3}=\frac{x-4}{4}\\ \Rightarrow & 12x-76=x^2-4x-3x+12\\ \Rightarrow & x^2-19x+88=0\\ \Rightarrow & x^2-11x-8x+88=0\\ \Rightarrow & (x-8)(x-11)=0\\ \Rightarrow & x=8\text{ or }x=11.\end{array}$

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8.4 AREAS OF SIMILAR TRIANGLS :

Statement: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Given: $$ Two triangles $ABC$ and $PQR$ such that $\mathrm{△}ABC\sim \mathrm{△}PQR$ [Shown in the figure]

To Prove : $\frac{ar(ABC)}{ar(PQR)}=(\frac{AB}{PQ}{)}^2=(\frac{BC}{QR}{)}^2=(\frac{CA}{RP}{)}^2$

Construction: Draw altitudes $AM$ and $PN$ of the triangle $ABC$ an $PQR$.

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Proof :

$\begin{array}{rl}& ar(ABC)=\frac{1}{2}BC\times AM\\ & \text{ And }ar(PQT)=\frac{1}{2}QR\times PN\\ & \text{ So, }\frac{ar(ABC)}{ar(PQR)}=\frac{\frac{1}{2}BC\times AM}{\frac{1}{2}QR\times PN}=\frac{BC\times AM}{QR\times PN}\end{array}$ …(i)

Now, in $\mathrm{△}ABM$ and $\mathrm{△}PQN$,

And $\mathrm{\angle }B=\mathrm{\angle }Q$ [As $\mathrm{△}ABC\sim \mathrm{△}PQR$ ]

$\mathrm{\angle }M=\mathrm{\angle }N$ $[{90}^0.each]$

So, $\mathrm{△}ABM\sim \mathrm{△}PQN$ [AA similarity criterion]

Therefore, $\frac{AM}{PN}=\frac{AB}{PQ}$ …(ii)

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Also, $\mathrm{△}ABC\sim \mathrm{△}PQR$ [given]

So, $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}$ …(iii)

Therefore,

$\frac{ar(ABC)}{ar(PQR)}=\frac{BC}{QR}\times \frac{AB}{PQ}$ $[\text{ From (i) and (ii)}]$

$=\frac{AB}{PQ}\times \frac{AB}{PQ}$ $[\text{ From (iii)}]$

$=(\frac{AB}{PQ}{)}^2$

$\text{ Now using (iii), we get }$ $\frac{ar(\mathrm{△}ABC)}{ar(\mathrm{△}PQR)}=(\frac{AB}{PQ}{)}^2=(\frac{BC}{QR}{)}^2=(\frac{CA}{RP}{)}^2$

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8.4 (a) Properties of Areas of Similar Triangles :

(i) The areas of two similar triangles are in the ratio of the squares of corresponding altitudes.

(ii) The areas of two similar triangles are in the ratio of the squares of the corresponding medians.

(iii) The area of two similar triangles are in the ratio of the squares of the corresponding angle bisector segments.

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Ex. 8 Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on this diagonals. $$ [CBSE - 2001]

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Sol. Given : A square $ABCD$. Equilateral triangles $\mathrm{△}BCE$ and $\mathrm{△}ACF$ have been described on side $BC$ and diagonals $AC$ respectively.

To prove : Area $(\mathrm{△}BCE)=\frac{1}{2}$. Area $(\mathrm{△}ACF)$

Proof : Since $\mathrm{△}BCE$ and $\mathrm{△}ACF$ are equilateral. Therefore, they are equiangular (each angle being equal to ${60}^{\circ }$ ) and hence $\mathrm{△}BCE\sim \mathrm{△}ACF$.

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$\begin{array}{rl}& \Rightarrow \frac{\text{ Area }(\mathrm{△}BCE)}{\text{ Area }(\mathrm{△}ACF)}=\frac{B{C}^2}{A{C}^2}\\ & \Rightarrow \frac{\text{ Area }(\mathrm{△}BCE)}{\text{ Area }(\mathrm{△}ACF)}=\frac{B{C}^2}{(\sqrt{2}BC{)}^2}=\frac{1}{2}\left[\begin{array}{c}\because ABCD\text{ is a square }\\ \because \text{ Diagonal }=\sqrt{2}(\text{ side })\\ \Rightarrow AC=\sqrt{2}BC\end{array}\right]\\ & \Rightarrow \frac{\text{ Area }(\mathrm{△}BCE)}{\text{ Area }(\mathrm{△}ACF)}=\frac{1}{2}\text{ Hence Proved. }\end{array}$

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8.5 PYTHAGOREOUS THEOREM :

Statement : In a right triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides.

Given : $A$ right triangle $ABC$, right angled at $B$.

To prove : $A{C}^2=A{B}^2+B{C}^2$

Construction: $BD\perp AC$

Proof : $\mathrm{\Delta }ADB$ & $\mathrm{\Delta }ABC$

$\angle DAB=\angle CAB$ $$ [Common]

$\angle BDA=\angle CBA$ $$ [${90}^{\circ }$ each]

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So, $∆ADB$ ~ $∆ABC$ $$ [By AA similarity]

$\frac{AD}{AB}=\frac{AB}{AC}$ $$ [Sides are proportional]

or, $AD.AC=A{B}^2$ …..(i)

Similarly $∆BDC∆ABC$

So, $\frac{CD}{BC}=\frac{BC}{AC}$

or $CD.AC=B{C}^2$ …..(ii)

Adding (i) and (ii),

$AD.AC+CD.AC=A{B}^2+B{C}^2$

or, $AC(AD+CD)=A{B}^2+B{C}^2$

or $AC.AC=A{B}^2+B{C}^2$

or, $A{C}^2=A{B}^2+B{C}^2$ Hence Proved.

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8.5 (a) Converse of Pythagoreans Theorem :

Statement : In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle.

Given : $$ A triangle $ABC$ such that $A{C}^2=A{B}^2+B{C}^2$

Construction : Construct a triangle DEF such that $DE=AB,EF=BC$ and $\mathrm{\angle }E={90}^{\circ }$

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Proof :

In order to prove that $\mathrm{\angle }B=90.{}^0$, it is sufficient to show $\mathrm{△}ABC\sim \mathrm{△}DEF$. For this we proceed as follows Since $\mathrm{\Delta }$ DEF is a right - angled triangle with right angle at $E$. Therefore, by Pythagoras theorem, we have

$\begin{array}{cc}D{F}^2=D{E}^2+E{F}^2& \\ \Rightarrow D{F}^2=A{B}^2+B{C}^2& [\therefore DE=AB\text{ and }EF=BC\text{ (By construction)] }\\ \Rightarrow D{F}^{2}3=A{C}^2& [\therefore A{B}^2+B{C}^2=A{C}^2\text{ (Given) }]\\ \Rightarrow DF=AC\dots ..(i)& \\ \text{ Thus, in }\mathrm{△}ABC\text{ and }\mathrm{△}DEF\text{, we have }& \\ AB=DE,BC=EF& \text{ [By construction] }\\ \text{ And }AC=DF& \text{ [From equation (i)] }\\ \therefore \mathrm{\Delta }ABC\cong \mathrm{\Delta }DEF& \text{ [By SSS criteria of congruency] }\\ \Rightarrow \mathrm{\angle }B=\mathrm{\angle }E={90}^0& \\ \text{ Hence, }\mathrm{△}ABC\text{ is a right triangle, right angled at }B.\end{array}$

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8.5 (b) Some Results Deduced From Pythagoreans Theorem :

(i) In the given figure $\mathrm{△}ABC$ is an obtuse triangle, obtuse angled at $B$. If $AD\perp CD$, then $A{C}^2=A{B}^2+B{C}^2+2\mathbf{B}\mathbf{C}\cdot \mathbf{B}\mathbf{C}$

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(ii) In the given figure, if $\mathrm{\angle }B$ of $\mathrm{△}ABC$ is an acute angle and $\mathbf{A}\mathbf{D}\perp \mathbf{B}\mathbf{C}$, then ${\mathbf{A}\mathbf{C}}^2={\mathbf{A}\mathbf{B}}^2+{\mathbf{B}\mathbf{C}}^2-\mathbf{2}\mathbf{B}\mathbf{C}\cdot \mathbf{B}\mathbf{D}$

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(iii) In any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side.

(iv) Three times the sum of the squares of the sides of a triangle is equal to four times the sum of the squares o the medians of the triangle.

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Ex. 9 In a $\mathrm{△}ABC,AB=BC=CA=2a$ and $AD\perp BC$. Prove that

(i) $AD=a\sqrt{3}$

(ii) area $(\mathrm{△}ABC)=\sqrt{3}{a}^2$

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Sol. (i) Here, $AD\perp BC$.

Clearly, $\mathrm{△}ABC$ is an equilateral triangle.

Thus, in $\mathrm{△}ABD$ and $\mathrm{△}ACD$

$AD=AD$

$\mathrm{\angle }ADB=\mathrm{\angle }ADC$

And $AB=AC$

$\therefore$ by RHS congruency condition

$\mathrm{△}ABD\cong \mathrm{△}ACD$

$\Rightarrow BD=DC=a$

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Now, $\mathrm{△}ABD$ is a right angled triangle

$\therefore AD=\sqrt{A{B}^2-B{D}^2}$ [Using Pythagoreans Theorem]

$AD=\sqrt{4a^2-a^2}=\sqrt{3}a$ or $a\sqrt{3}$

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(ii)

$\begin{array}{rl}\text{ Area }(\mathrm{△}ABC)=& \frac{1}{2}\times BC\times AD\\ & =\frac{1}{2}\times 2a\times a\sqrt{3}\\ & =a^2\sqrt{3}\end{array}$

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Ex. 10 $BL$ and $Cm$ are medians of $\mathrm{△}ABC$ right angled at $A$. Prove that $4(B{L}^2+C{M}^2)=5B{C}^2$

[CBSE-2006]

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Sol. In $\mathrm{△}BAL$

$B{L}^2=A{L}^2+A{B}^2$ …(i) [Using Pythagoreans theorem]

and In $\mathrm{△}CAM$

$C{M}^2=A{M}^2+A{C}^2$ …(ii) [Using Pythagoreans theorem]

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Adding (1) and (2) and then multiplying by 4 , we get

$4(B{L}^2+C{M}^2)=4(A{L}^2+A{B}^2+A{M}^2+A{C}^2)$

$=4A{L}^2+A{M}^2+(A{B}^2+A{C}^2)[\therefore \mathrm{\Delta }ABC$ is a right triangle $]$

$=4(A{L}^2+A{M}^2+B{C}^2)$

$=4(M{L}^2+B{C}^2)$ $[\therefore \mathrm{\Delta }LAM$ is a right triangle]

$=4M{L}^2+4B{C}^2$

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[A line joining mid-points of two sides is parallel to third side and is equal to half of it, $ML=BC/2$ ]

$=B{C}^2+4B{C}^2=5B{C}^2$

Hence proved.

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Ex. 11 In the given figure, $BC\perp AB,AE\perp AB$ and $DE\perp AC$. Prove that $DE.BC=AD.AB$.

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Sol. In $\mathrm{△}ABC$ and $\mathrm{△}EDA$,

We have

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$\begin{array}{rl}& \mathrm{\angle }ABC=\mathrm{△}ADE\\ & \mathrm{\angle }ACB=\mathrm{\angle }EAD\\ \therefore & ByAA\text{ Similarity }\\ & \mathrm{\Delta }ABC\sim \mathrm{\Delta }EDA\\ \Rightarrow & \frac{BC}{AB}=\frac{AD}{DE}\\ \Rightarrow & DE\cdot BC=AD\cdot AB.\end{array}$

Hence Proved.

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Ex. 12 $O$ is any point inside a rectangle $ABCD$ (shown in the figure). Prove that $O{B}^2+O{D}^2=O{A}^2+O{C}^2$

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Sol. Through $O$, draw $PQ||BC$ so that $P$ lies on $A$ and $Q$ lies on DC. [CBSE - 2006]

Now, $PQ||BC$

Therefore,

$PQ\perp AB$ and $PQ\perp DC[\mathrm{\angle }B={90}^{\circ }.$ and $.\mathrm{\angle }C={90}^{\circ }]$

So, $\mathrm{\angle }BPQ={90}^{\circ }$ and $\mathrm{\angle }CQP={90}^{\circ }$

Therefore, $BPQC$ and $APQD$ are both rectangles.

Now, from $\mathrm{△}OPB$,

$O{B}^2=B{P}^2+O{P}^2$ …(i)

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Similarly, from $\mathrm{△}ODQ$,

From $\mathrm{\Delta }OQC$, we have

$O{D}^2=O{Q}^2+D{Q}^2$ …(ii)

$O{C}^2=O{Q}^2+C{Q}^2$ …(iii)

And form $\mathrm{△}OAP$, we have

$O{A}^2=A{P}^2+O{P}^2$ …(iv)

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Adding (i) and (ii)

$\begin{array}{rl}& O{B}^2+O{D}^2=B{P}^2+O{P}^2+O{Q}^2+D{Q}^2\\ & =C{Q}^2+O{P}^2+O{Q}^2+A{P}^2\\ & [AsBP=CQ\text{ and }DQ=AP]\\ & =C{Q}^2+O{Q}^2+O{P}^2+A{P}^2\\ & =O{C}^2+O{A}^2[\text{ From (iii) and (iv) }]\end{array}$

Hence Proved.

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Ex. 13 $ABC$ is a right triangle, right-angled at $C$. Let $BC=a,CAb,AB=c$ and let $p$ be the length of perpendicular

form $C$ on $AB$, prove that

(i) $cp=ab$ (ii) $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$

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Sol. Let $CD\perp AB$. Then $CD=p$

$\therefore$ Area of $\mathrm{△}ABC=\frac{1}{2}$ (Base $\times$ height $)$

$=\frac{1}{2}(AB\times CD)=\frac{1}{2}cp$

Also,

Area of $\mathrm{△}ABC=\frac{1}{2}(BC\times AC)=\frac{1}{2}ab$

$\therefore \frac{1}{2}cp=\frac{1}{2}ab$

$\Rightarrow CP=AB$.

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(ii) Since $\mathrm{△}ABC$ is a right triangle, right angled at $C$.

$\therefore A{B}^2=B{C}^2+A{C}^2$

$\Rightarrow c^2=a..+b^2$

$\Rightarrow (\frac{ab}{p}{)}^2=a^2+b^2[\because cp=ab\Rightarrow c=\frac{ab}{p}]$

$\Rightarrow \frac{a^2{b}^2}{p^2}=a^2+b^2\Rightarrow \frac{1}{p^2}=\frac{1}{b^2}+\frac{1}{a^2}\Rightarrow \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$

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Ex. 14 In an equilateral triangle $ABC$, the side $B$ is trisected at $D$. Prove that $9A^2=7A{B}^2$.

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Sol. $ABC$ be can equilateral triangle and $D$ be point on $BC$ such that [CBSE - 2005]

$BC=\frac{1}{3}BC$ (Given)

Draw $AE\perp BC$, Join $AD$.

$BE=EC$ (Altitude drown from any vertex of an equilateral triangle bisects the opposite side)

So, $BE=EC=\frac{BC}{2}$

In $\mathrm{△}ABC$

$\begin{array}{rl}& A{B}^2=A{E}^2+E^2\dots (i)\\ & A{D}^2=A{E}^2+E^2\dots (ii)\end{array}$

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From (i) and (ii)

$\begin{array}{rl}& A{B}^2=A{D}^2-E^2+E^2\\ & A{B}^2=A{D}^2-\frac{B{C}^2}{36}+\frac{B{C}^2}{4}(\therefore BD+DE=\frac{BC}{2}\Rightarrow \frac{BC}{3}+DE=\frac{BC}{2}\Rightarrow DE=\frac{BC}{6})\end{array}$

$A{B}^2+\frac{B{C}^2}{36}-\frac{B{C}^2}{4}=A{D}^2(\therefore EB=\frac{BC}{2})$

$A{B}^2+\frac{A{B}^2}{36}-\frac{A{B}^2}{4}=A{D}^2(\therefore AB=BC)$

$\frac{36A{B}^2+A{B}^2-9A{B}^2}{36}=A{D}^2\Rightarrow \frac{28AB}{36}=A{D}^2$

$7A{B}^2=9A{D}^2$

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DAILY PRACTIVE PROBLEMS 8

OBJECTIVE DPP - 8.1

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1. The perimeters of two similar triangles are $25cm$ and $15cm$ respectively. If one side of first triangle is 9 $cm$, then the corresponding side of the other triangle is

(A) $6.2cm$ $$

(B) $3.4cm$ $$

(C) $5.4cm$ $$

(D) $8.4cm$

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2. In the following figure, $AE\perp BC,D$ is the mid point of $BC$, hen $x$ is equal to

(A) $\frac{1}{a}[b^2-d^2-\frac{a^2}{4}]$ $$

(B) $\frac{h+d}{3}$ $$

(C) $\frac{c+d-h}{2}$ $$

(D) $\frac{a^2+b^2+d^2-c^2}{4}$

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3. Two triangles $ABC$ and $PQR$ are similar, if $BC:CA:AB=1:2:3$, then $\frac{QR}{PR}$ is

(A) $\frac{2}{3}$ $$

(B) $\frac{1}{2}$ $$

(C) $\frac{1}{\sqrt{2}}$ $$

(D) $\frac{2}{3}$

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4. In a triangle $ABC$, if angle $B={90}^0$ and $D$ is the point in $BC$ such that $BD=2DC$, then

(A) $A{C}^2=A{D}^2+3C{D}^2(B)$ $$

(B) $A{C}^2=A{D}^2+5C{D}^2$ $$

(C) $A{C}^2=A{D}^2+7C{D}^2$ $$

(D) $A{C}^2=A{B}^2+5B{D}^2$

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5. $P$ and $Q$ are the mid points of the sides $AB$ and $BC$ respectively of the triangle $ABC$, right-angled at $B$, then

(A) $A{Q}^2+C{P}^2=A{C}^2$ $$

(B) $A{Q}^2+C{P}^2=\frac{4}{5}A{C}^2$ $$

(C) $A{Q}^2+C{P}^2=\frac{5}{4}A{C}^2$ $$

(D) $A{Q}^2+C{P}^2=\frac{3}{5}A{C}^3$

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6. In a $\mathrm{△}ABC,AD$ is the bisector of $\mathrm{\angle }A$, meeting side $BC$ at $D$.

If $AB=10cm,AC=6cm,BC=12cm$, find $BD$.

(A) 3.3 $$

(B) 18 $$

(C) 7.5 $$

(D) 1.33

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7. In a triangle $ABC$, a straight line parallel to $BC$ intersects $AB$ and $AC$ at point $D$ and $E$ respectively. If the area of $ADE$ is one-fifth of the area of $ABC$ and $BC=10cm$, then $DE$ equals

(A) $2cm$ $$

(B) $2\sqrt{5}cm$ $$

(C) $4cm$ $$

(D) $4\sqrt{5}cm$

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8. $ABC$ is a right-angle triangle, right angled at $A$. A circle is inscribed in it. The lengths of the two sides containing the right angle are $6cm$ and $8cm$, then radius of the circle is

(A) $3cm$ $$

(B) $2cm$ $$

(C) $4cm$ $$

(D) $8cm$

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SUBJECTIVE DPP - 8.2

1. Given $\mathrm{\angle }GHE=\mathrm{\angle }DFE={90}^{\circ },DH=8,DF=12,DG=3x-1$ and $DE=4x+2$.

Find the lengths of segments DG and DE.

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Sol. 1. $20$ $unit$ & $30$ $unit$ $$

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2. In the given figure, $DE$ is parallel to the base $BC$ of triangle $ABC$ and $AD:DB=5:3$. Find the ratio :-

(i) $\frac{AD}{AB}$ $$

(ii) $\frac{\text{ Area of }\mathrm{△}DEF}{\text{ Area of }\mathrm{△}CFB}$

[CBSE - 2000]

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Sol.2. (i) $\frac{5}{8}$ (ii) $\frac{25}{64}$ $$

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3. In Figure, $\mathrm{△}ABC$ is a right-angled triangle, where $\mathrm{\angle }ACB={90}^{\circ }$. The external bisector $BD$ of $\mathrm{\angle }ABC$ meets $AC$ produced at $D$. If $AB=17cm$ and $BC=8cm$, find the $AC$ and $BD$.

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Sol.3. $15cm.,\frac{8\sqrt{34}}{3}cm$. $$

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4. In figure, $\mathrm{\angle }QPS=\mathrm{\angle }RPT$ and $\mathrm{\angle }PST=\mathrm{\angle }PQR$. Prove that $\mathrm{△}PST\sim \mathrm{△}PQR$ and hence find the ratio $ST:PT$, if $PR:R=4:5$.

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Sol.4. $5:4$ $$

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5. In the figure, $PQRS$ is a parallelogram with $PQ=16cm$ and $QR=10cm$. $L$ is a point on $PR$ such that $RL$ : $LP=2$ : 3. QL produced meets RS at $M$ and PS produced at $N$.

Find the lengths of PN and RM.

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Sol.5. $PN=15cm,RM=10.67cm$. $$

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6. In $\mathrm{△}ABC,D$ and $E$ are points on $AB$ and $AC$ respectively such that $DE||BC$. If $AD=2.4cm,AE=3.2cm$, $DE=2cm$ and $BC=5cm$, find $BD$ and $CE$.

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Sol.6. $DB=3.6cm,CE=4.8cm$ $$

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7. In a triangle $PQR,L$ an $DM$ are two points on the base $QR$, such that $\mathrm{△}:PQ=\mathrm{\angle }QRP$ and $\mathrm{\angle }RPM=\mathrm{\angle }RQP$. Prove that :

(i) $\mathrm{\Delta }PQL\sim \mathrm{\Delta }RPM$

(ii) $QL\times RM=PL\times PM$

(iii) $P{Q}^2=QR\times QL$

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8. In figure, $\mathrm{\angle }BAC={90}^0,AD\perp BC$. prove that $A{B}^2=B{D}^2-C{D}^2$.

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9. In figure, $\mathrm{\angle }ACB={90}^0,CD\perp AB$ prove that $C{D}^2=BD\cdot AD$.

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10. In a right triangle, prove that the square on the hypotenuse is equal to sum of the squares on the other two sides.

Using the above result, prove the following:

In figure $PQR$ is a right triangle, right angled at $Q$. If $QS=SR$, show that $P{R}^2=4P{S}^2-3P{Q}^2$.

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11. In $\mathrm{△}ABC,\mathrm{\angle }ABC={135}^0$. Prove that $A{C}^2=A{B}^2+B{C}^2+4ar(\mathrm{△}ABC)$.

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12. In figure, $ABC$ and $DBC$ are two right triangles with the common hypotenuse $BC$ and with their sides $AC$ and $DB$ intersecting at $P$. Prove that $AP\times PC=DP\times PB$.

[CBSE- 2000]

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13. Any point $O$, inside $\mathrm{△}ABC$, in joined to its vertices. From a point $D$ on $AO,DE$ is drawn so that $DE||AB$ and $EF||BC$ as shown in figure. Prove that $DF||AC$.