Solution

A median of a triangle divides it into two triangles of equal area.

Hence, the correct option is (A).

Solution

In figure (d), we find two polygons (PQRA and BQRS) on the same base and between the same parallels.

Hence, the correct option is (D).

Solution

According to the question,

$ABCD$ is a rectangle and $E, F, G$ and $H$ are the mid-point of the sides $AB, BC, CD$ and $DA$ respectively. The figure formed is rhombus hose area:

$ \begin{aligned} & =\dfrac{1}{2} \times E G \times F H \\ & =\dfrac{1}{2} \times 6 ~cm \times 8 ~cm \\ & =24 ~cm^{2} \end{aligned} $

Hence, the correct option is (D).

Solution

Area of parallelogram $=$ Base $\times$ Corresponding altitude

$=A B \times D L=D C \times D L$

[Since, AB = DC (opposite side of a parallelogram)]

Hence, the correct option is (C).

Solution

If parallelogram $ABCD$ and rectangle $ABEF$ are of equal area then perimeter of $ABCD>$ Perimeter of ABEM because:

As we know that, the perpendicular distance between two parallel sides of a parallelogram is always less than the length of the other parallel sides.

$BE<BC$ and $AM<AD$.

Hence, the correct option is (C).

Solution

We know that, median of a triangle divides it into two triangle of equal area.

So, area $(\triangle A D E)=area(\triangle B D E)$

$area(\triangle A E F)=area(\triangle E F C)$

Now, $AE$ is the diagonal of a parallelogram ADEF. That is divides it into two triangles of equal area.

So, $area(\triangle A D E)=area(\triangle A F E)$

Now, from equation (I), (II), and (III), get:

$area(\triangle A D E)=area(\triangle B D E)=area(\triangle A E E)=area(\triangle E F C)$

Hence, $area(\triangle A D E F)=\dfrac{1}{2} area(\triangle A B C)$

Therefore, the correct option is (A).

Solution

As we know that parallelogram on the same or equal bases and between the same parallels are equal in area.

So, the ratio of these area is $1: 1$.

Hence, the correct option is (B).

Solution

The quadrilateral $ABCD$ need not be any of rectangle, rhombus and parallelogram because if quadrilateral $ABCD$ is a square then its diagonal $AC$ also divides it into two parts which are equal in area.

Hence, the correct option is (D).

Solution

As we know that, if a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram. Therefore, the ratio of the area of the triangle to the area of parallelogram is 1:2.

Hence, the correct option is (B).

Solution

Given:

$ABCD$ is a trapezium with parallel sides such that $AB || DC$ and $AB=a ~cm$ and $DC=b ~cm . E$ and $F$ are the mid-points of the non-parallel sides that are $AD$ and $BC$. So,

$ E F=\dfrac{1}{2}(a+b) $

$ABEF$ and EFCD are also trapezium.

$ \begin{aligned} & area(A B E F)=\dfrac{1}{2}[\dfrac{1}{2}(a+b)+a] \times h=\dfrac{h}{4}(3 a+b) \\ & area(E F C D)=\dfrac{1}{2}[b+\dfrac{1}{2}(a+b)] \times h=\dfrac{h}{4}(a+3 b) \end{aligned} $

So,

$\dfrac{area(A B E F)}{area(E F C D)}=\dfrac{\dfrac{h}{4}(3 a+b)}{\dfrac{h}{4}(a+3 b)}=\dfrac{3 a+b}{a+3 b}$

So, the required ratio is $(3 a+b):(a+3 b)$.

Hence, the correct option is (B).

Solution

Given in the question, $ABCD$ is a parallelogram and $X$ is the mid-point of $AB$.

So, $area(A B C D)=area(A X C D)+area(\triangle X B C)$

Now, diagonal $AC$ of a parallelogram divides it into two triangles of equal area.

area $(A B C D)=2 area(\triangle A B C)$

Similarly, $X$ is the mid-point of $AB$, So,

area $(\triangle C X B)=\dfrac{1}{2}$ area $(\triangle A B C) \quad \ldots$ (III) [Median divides the triangle in two triangles of equal area]

2area $(\triangle A B C)=24+\dfrac{1}{2} area(\triangle A B C) \quad$ [By using equation (I), (II) and (III)]

Now, 2area $(\triangle A B C)-\dfrac{1}{2} area(\triangle A B C)=24$

$\dfrac{3}{2} area(\triangle A B C)=24$

Therefore, $area(\triangle A B C)=\dfrac{2 \times 24}{3}=16 ~cm^{2}$.

Solution

Given: A is any point on PQ. Since, $PA<PQ$

Now, area of triangle PQR is:

area $(\triangle P Q R)=\dfrac{1}{2} \times$ base $\times$ height

So, $area(\triangle P Q R)=\dfrac{1}{2} \times P Q \times Q R=\dfrac{1}{2} \times 12 \times 5=30 ~cm^{2}[PQRS$ is a rectangle, $RQ=SP=5 ~cm]$

As $PA<PQ$

So, area $(\triangle P A S)<area(\triangle P Q R)$

Or area $(\triangle P A S)<30 ~cm^{2} \quad[area(\triangle P Q R)=30 ~cm^{2}]$

Hence, the given statement is false.

Solution

Given: PQRS is a parallelogram.

As we know that diagonal of a parallelogram divides parallelogram into two triangles of equal area.

So,

$ \begin{aligned} area(\triangle Q R S) & =\dfrac{1}{2} area(P Q R S) \\ & =\dfrac{1}{2} \times 180=90 ~cm^{2} \end{aligned} $

Now, $A$ is any point on SQ. So, area $(\triangle A S R)<area(\triangle Q R S)$

Therefore, $area(\triangle A S R)<90 ~cm^{2}$

Hence, the given statement is false.

Solution

Given: $\triangle ABC$ and $\triangle BDE$ are two equilateral triangles.

Suppose that each sides of triangle $ABC$ be $x$.

Similarly, D is the mid-point of BC. So, each side of triangle BDE is $\dfrac{x}{2}$.

Now,

$\dfrac{area(\triangle B D E)}{area(\triangle A B C)}=\dfrac{\dfrac{\sqrt{3}}{4} \times(\dfrac{x}{2})^{2}}{\dfrac{\sqrt{3}}{4} \times x^{2}}=\dfrac{x^{2}}{4 x^{2}}=\dfrac{1}{4}$

Therefore, area $(\triangle B D E)=\dfrac{1}{4} area(\triangle A B C)$.

Hence, the given statement is true.

Solution

As triangle DPC and parallelogram ABCD are on same base DC and between the same parallels $AB$ and $DC$. So,

$$ \begin{equation*} area(\triangle D P C)=\dfrac{1}{2} area(A B C D) \tag{I} \end{equation*} $$

Now,

$\dfrac{area(E F G D)}{area(A B C D)}=\dfrac{D G \times h}{D C \times h}=\dfrac{D G}{2 D G}=\dfrac{1}{2}(G$ is the mid-point of $DC)$

Implies that, area $(E F G D)=\dfrac{1}{2} area(A B C D)$

So, area $(D P C)=area(E F G D) \quad[$ From equation $(I)]$

Hence, the given statement is false.

Solution

Given: PSDA is a parallelogram. Points $Q$ and $R$ are taken on $PS$ such that $PQ=QR=RS$ and PA $||$ QB $||$ RC.

To prove that ar $(PQE)=ar(CFD)$.

Proof: $PS=AD \quad$ [opposite angle of a parallelogram]

$\dfrac{1}{3} P S=\dfrac{1}{3} A D$

$PQ=CD$

Similarly, PS $|| AD$ and $QB$ cut them. So,

$\angle P Q E=\angle C B E \quad$ [Alternate angles]…(II)

Again, $QB || RC$ and $AD$ cut them, $\angle Q B D=\angle R C D$ [Corresponding angle]

So, $\angle P Q E=\angle F C D \quad \ldots$ (IV) [From (II) and (III), $\angle C B E$ and $\angle Q B D$ are same and $\angle R C D$ and $\angle F C D$ are same]

Now, in triangle $PQE$ and triangle $CFD$,

$\angle P Q E=\angle C D F \quad$ [Alternate angle]

$PQ=CD \quad$ [From equation (I)]

$\angle Q P E=\angle F C D \quad[$ From equation (IV)]

$\triangle P Q E \cong \triangle C F D \quad$ [By ASA congruence rule]

Hence, $ar(\triangle P Q E)=ar(\triangle C F D)$. [Congruence triangle are equal in area]

Solution

Prove that ar $(LZY)=ar(MZYX)$

Proof: As $\triangle L X Z$ and $\triangle X M Z$ are on the same base and between the same parallels LM and XZ.

$ar(\triangle L X Z)=ar(\triangle X M Z)$

Now, adding $ar(\triangle X Y Z)$ to both sides of (I), get:

$ \begin{aligned} ar(\triangle L X Z)+ar(\triangle X Y Z) & =ar(\triangle X M Z)+ar(\triangle X Y Z) \\ ar(\triangle L Z Y) & =ar(M Z Y X) \end{aligned} $

Solution

(i) As we know that parallelogram on the same base and between the same parallels are equal in area.

$ar(A B E F)=ar(A B C D)$

Hence, $ar(A B E F)=ar(A B C D)=90 ~cm^{2}$.

(ii) $\quad ar(\triangle A B D)=\dfrac{1}{2} ar(A B C D)$ [A diagonal of a parallelogram divides the parallelogram in two triangle of equal area]

$=\dfrac{1}{2} \times 90 ~cm^{2}=45 ~cm^{2}$

(iii) $ar(\triangle B E F)=\dfrac{1}{2} ar(A B E F)=\dfrac{1}{2} \times 90 ~cm^{2}=45 ~cm^{2}$

Solution

Given in triangle $ABC, D$ is the mid-point of $AB$ and $P$ is any point on $BC$.

$CQ || PD$ means $AB$ in $Q$.

To prove that $ar(\triangle B P Q)=\dfrac{1}{2} ar(\triangle A B C)$

Construction: Join PQ and CD.

Proof:

As we know that median of a triangle divides it into two triangles of equal area. So,

$ar(\triangle B C D)=\dfrac{1}{2} ar(\triangle A B C)$

Also, we know that triangles on the same base and between the same parallels are equal in area. So,

$ar(\triangle D P Q)=ar(\triangle D P C) \quad[$ Triangle DPQ and DPC are on the same base DP and between the same parallels DP and CQ]

$ar(\triangle D P Q)+ar(\triangle D P B)=ar(\triangle D P C)+ar(D P B)$

Hence, $ar(\triangle B P Q)=ar(\triangle B C D)=\dfrac{1}{2} ar(\triangle A B C)$.

Solution

Given: $ABCD$ is a square. $E$ and $F$ are respectively the midpoints of $BC$ and $CD$. Also, $R$ is the mid-point of EF.

To prove that $ar(\triangle A E R)=ar(\triangle A F R)$

Construction: Draw $AN \perp EF$

Proof:

$ar(\triangle A E R)=\dfrac{1}{2} \times$ Base $\times$ Height

$ \begin{aligned} & =\dfrac{1}{2} \times E R \times A N \\ & =\dfrac{1}{2} \times F R \times A N \quad[R \text{ is the mid-point of EF so ER }=FR] \\ & =ar(\triangle A F R) \end{aligned} $

Solution

Given: $O$ is any point on the diagonal PR of a parallelogram PQRS.

To prove that ar $(PSO)=ar(PQO)$.

Construction: Join SQ which intersects PR at B.

Proof: B is the mid-point of SQ because diagonal of a parallelogram bisect each other.

See the above figure, $PB$ is a median of $\triangle Q P S$ and as we know that a median of a triangle divides it into two triangle of equal area.

$ar(\triangle B P Q)=ar(\triangle B P S)$

Similarly, OB is the median of $\triangle O S Q$.

$ar(\triangle O B Q)=ar(O B S)$

Now, adding equation (I) and (II), get:

$ar(\triangle B P Q)+ar(\triangle O B Q)=ar(\triangle B P S)+ar(\triangle O B S)$

$ar(\triangle P Q O)=ar(\triangle P S O)$

Solution

Given: $ABCD$ is a parallelogram in which $BC$ is produced to $E$ such that $CE=BC . C$ is the mid-point $BE$ and $ar(\triangle D F B)=3 ~cm^{2}$.

In triangle $ADF$ and triangle $EFC$,

$\angle D A F=\angle C E F \quad$ [Alternate interior angle]

$AD=CE \quad[AD=BC=CE]$

$\angle A D F=\angle F C E \quad$ [Alternate interior angle]

So, $\triangle A D F \cong \triangle E C F \quad$ [By SAS rule of congruence]

Now, $\triangle A D F \cong \triangle E C F \quad$ [By SAS rule of congruence] $D F=C F$ $[CPCT]$

$As BF$ is the median of triangle $BCD$.

$ar(\triangle B D F)=\dfrac{1}{2} ar(B C D) \quad \ldots$ (I) [Median divides a triangle into two triangle of equal area]

As we know that a triangle and parallelogram are on the same base and between the same parallels then the area of the triangles is equal to half the area of the parallelogram.

$$ \begin{align*} & ar(\triangle B C D)=\dfrac{1}{2} ar(A B C D) \quad \ldots(II) \tag{II}\\ & ar(\triangle B D F)=\dfrac{1}{2}{\dfrac{1}{2} ar(A B C D)} \quad \text{ [By equation (I)] } \\ & 3 ~cm^{2}=\dfrac{1}{4} ar(A B C D) \\ & ar(A B C D)=12 ~cm^{2} \end{align*} $$

Hence, the area of the parallelogram is $12 ~cm^{2}$.

Solution

To prove that ar $(ABCD)=ar(APQD)$

Proof: $A S A B || D C$ and $A B || D Q$

In triangle CLQ and triangle BLP,

$ \angle Q C L=\angle L B P \quad \text{ [Alternate angles] } $

$CL=LP \quad[L$ is the mid-point og $BC]$

$\angle C L Q=\angle B L P \quad$ [Vertical opposite angles]

$\triangle C L Q \cong \triangle B L P \quad$ [By ASA congruence rule]

So, $ar(\triangle C L Q)=ar(\triangle B L P) \ldots$ (I) [Congruent triangles are equal in area]

Now, adding $ar(A P L C D)$ both side in above equation, get:

$ \begin{aligned} ar(\triangle C L Q)+ar(A P L C D) & =ar(\triangle B L P)+ar(A P L C D) \\ ar(\triangle A P Q D) & =ar(A B C D) \end{aligned} $

Hence, proved.

Solution

According to the question, a quadrilateral $ABCD$ in which the mid-point of the sides of it are joined in order of form parallelogram PQRS.

To Prove that $ar(P Q R S)=\dfrac{1}{2} ar(A B C D)$

Construction: Join BD and draw perpendicular from A and BD which interest SR and BD at $X$ and $Y$ respectively.

Proof: In triangle $ABD, S$ and $R$ are the mid-points of sides $AB$ and $AD$ respectively. So, $S R || B D$

And: $A S X || B Y$

See the figure, $x$ is the mid-point of AY. So,

$A X=X Y$

And $S R=\dfrac{1}{2} B D \ldots$ (II)[mid-point theorem]

Now, $ar(\triangle A B D)=\dfrac{1}{2} \times B D \times A Y$ $ar(\triangle A S R)=\dfrac{1}{2} \times S R \times A X$

$ar(\triangle A S R)=\dfrac{1}{2} \times(\dfrac{1}{2} B D) \times(\dfrac{1}{2} A Y) \quad[$ Using equation (I) and (II)]

$ar(\triangle A S R)=\dfrac{1}{4} \times(\dfrac{1}{2} B D \times A Y)$

$ar(\triangle A S R)=\dfrac{1}{4} \times(\triangle A B D)$

Again, $ar(\triangle C P Q)=\dfrac{1}{4} ar(\triangle C B D)$

$ar(\triangle B P S)=\dfrac{1}{4} ar(\triangle B A C)$

$ar(\triangle D R Q)=\dfrac{1}{4} ar(D A C)$

Now, adding equation (III), (IV), (V) and (VI), get:

$ \begin{aligned} & ar(\triangle A S R)+ar(\triangle C P Q)+ar(B P S)+ar(\triangle D R Q) \\ & =\dfrac{1}{4} ar(\triangle A B D)+\dfrac{1}{4} ar(\triangle C B D)+\dfrac{1}{4} ar(\triangle A B C)+\dfrac{1}{4} ar(\triangle D A C) \\ & =\dfrac{1}{4}[ar(\triangle A B D)+ar(\triangle C B D)+ar(\triangle A B C)+ar(\triangle D A C)] \\ & =\dfrac{1}{4}[ar(A B C D)+ar(A B C D)] \\ & =\dfrac{1}{4} \times 2 ar(A B C D) \\ & =\dfrac{1}{2} ar(A B C D) \end{aligned} $

So, $ar(\triangle A S R)+ar(\triangle C P Q)+ar(\triangle B P S)+ar(\triangle D R Q)=\dfrac{1}{2} ar(A B C D)$

$ ar(A B C D)-ar(P Q R S)=\dfrac{1}{2} ar(A B C D) $

Now, $ar(P Q R S)=ar(A B C D)-\dfrac{1}{2} ar(A B C D)$

$ ar(P Q R S)=\dfrac{1}{2} ar(A B C D) $

Hence, proved.

Solution

Given in the question, $A$ point $E$ is taken on the side $BC$ of a parallelogram $ABCD . AE$ and $DC$ are produced to meet at $F$.

Prove that ar $(A D F)=ar(A B F C)$

Proof: $ABCD$ is a parallelogram and $AC$ divides it into two triangle of equal area.

$ar(\triangle A D C)=ar(\triangle A B C)$

So, $DC || AB$ and $CF || AB$

As we know that triangle on the same base and between the same parallels are equal in area. So,

$ar(\triangle A C F)=ar(\triangle B C F)$

Adding equation (I) and (II), get:

$ar(\triangle A D C)+ar(A C F)=ar(\triangle A B C)+ar(\triangle B C F)$

$ar(\triangle A D F)=ar(A B F C)$

Hence, proved.

Solution

Given: $ABCD$ is a parallelogram and diagonal interact at $O$, and draw a line PQ which intersects $AD$ and $BC$.

To prove that $PQ$ divides the parallelogram $ABCD$ into two parts of equal area that $ar(A B Q P)=ar(C D P Q)$.

Proof: $A C$ is a diagonal of the parallelogram $A B C D$.

$ar(\triangle \dfrac{1}{2} A C D)=\dfrac{1}{2} ar(A B C D)$

In triangle $AOP$ and triangle $COQ$, $AO=CO$ [Diagonals of a parallelogram bisect each other] $\angle A O P=\angle C O Q$ [Vertical opposite angles] $\angle O A P=\angle O C Q$ [Alternate angles, $AB | CD$ ] $\triangle A O P=\triangle C O Q$ [By ASA congruent rule]

Since, $ar(\triangle A O P)=ar(\triangle C O Q) \quad$ [Congruent area axiom]

Now, adding $ar(A O Q D)$ to both sides of (II), get:

$ar(\triangle A C D)=\dfrac{1}{2} ar(A B C D) \quad$ [From equation (I)]

Hence, $ar(A P Q D)=\dfrac{1}{2} ar(A B C D)$.

Solution

Given: The medians $BE$ and $CF$ of a triangle $ABC$ intersect at $G$ To prove that $ar(\triangle G B C)=ar(A F G E)$.

Proof: As median $CF$ divides a triangle into two triangle of equal area. So,

$ar(\triangle B C F)=ar(\triangle A C F)$

$ar(\triangle G B F)+ar(\triangle G B C)=ar(A F G E)+ar(\triangle G C E)$

Now, median $BE$ divides a triangle into two triangle of equal area. So,

$$ \begin{equation*} ar(\triangle G B F)+ar(A F G E)=ar(\triangle G C E)+ar(\triangle G B C) \tag{II} \end{equation*} $$

Now, subtracting (II) from (I), get:

$ \begin{aligned} ar(\triangle G B C)-ar(A F G E) & =ar(\triangle A F G E)-ar(\triangle G B C) \\ ar(\triangle G B C)+ar(\triangle G B C) & =ar(\triangle A F G E)+ar(\triangle A F G E) \\ 2 ar(\triangle G B C) & =2 ar(A F G E) \end{aligned} $

Hence, $ar(\triangle G B C)=ar(A F G E)$.

Solution

Given: $CD || AE$ and $CY || BA$

To prove that ar $(C B X)=$ ar $(A X Y)$.

Proof: As we know that triangle on the same base and between the same parallels are equal in area. So,

$ \begin{aligned} ar(\triangle A B C) & =ar(\triangle A B Y) \\ ar(\triangle C B X)+ar(\triangle A B X) & =ar(\triangle A B X)+ar(\triangle A X Y) \end{aligned} $

Hence, $ar(\triangle C B X)=ar(\triangle A X Y)$.

Solution

To prove that $ar(DCYX)=\dfrac{7}{9} ar(XYBA)$

Proof: In triangle MBY and triangle DCY,

$\angle 1=\angle 2 \quad$ [Vertically opposite angles]

$\angle 3=\angle 4 \quad[AB || DC$ and alternate angles are equal $]$

$BY=CY \quad[Y$ is the mid-point of $BC]$

$\triangle M B Y \cong \triangle D C Y \quad$ [By ASA congruent angle]

So, $MB=DC=30 ~cm \quad[CPCT]$

Now, $AM=AB+BM$ $=50 ~cm+30 ~cm$ $=80 ~cm$

In triangle $ADM$,

$X Y=\dfrac{1}{2} A M=\dfrac{1}{2} \times 80 ~cm=40 ~cm$

Now, $AB||XY|| DC$ and $X$ and $Y$ are the mid-points of $AD$ and $BC$, So, height of trapezium DCXY and XYBA are equal and assume the equal height be $h ~cm$.

$\dfrac{ar(D C Y X)}{ar(X Y B A)}=\dfrac{\dfrac{1}{2} \times(30+40) \times h}{\dfrac{1}{2} \times(30+50) \times h}=\dfrac{70}{90}=\dfrac{7}{9}$

Hence, $ar(D C Y X)=\dfrac{7}{9} ar(X Y B A)$.

Solution

Given: in triangle $ABC$ and $L$ and $M$ are the points on $AB$ and $AC$, respectively such that $LM$ || BC.

Prove that ar $(LOB)=ar(MOC)$

Proof: As we know that triangle on the same base and between the same parallels are equal in area.

$ar(\triangle L B M)=ar(\triangle L C M)$

[Triangle LBM and triangle L~CM are on the same base LM and between the same parallels LM and $BC]$

$\begin{aligned} ar(\triangle L B M) & =ar(\triangle L C M) \\ ar(\triangle L O M)+ar(\triangle L O B) & =ar(\triangle L O M)+ar(\triangle M O C)\end{aligned}$

Hence, $ar(\triangle L O B)=ar(\triangle M O C)$.

Solution

Given: $ABCDE$ is any pentagon and $BP || AC$ meets $DC$ produced at $P$ and $EQ || AD$ meets $CD$ produced at $Q$.

Prove that ar $(ABCDE)=ar(APQ)$

Proof: As we know that triangle on the same base and between the same parallels are equal in area.

$ar(\triangle A B C)=ar(\triangle A P C)$

$ar(\triangle A D E)=ar(\triangle A D Q)$

Now, adding equation (I) and (II), get:

$ar(\triangle A B C)+ar(\triangle A D E)=ar(\triangle A P C)+ar(\triangle A D Q)$

Now, adding $ar(\triangle A C D)$ to both side, get:

$ar(\triangle A B C)+ar(\triangle A D E)+ar(A C D)=ar(\triangle A P C)+ar(\triangle A D Q)+ar(A C D)$

Hence, $ar(A B C D E)=ar(\triangle A P Q)$.

Solution

Given: The median of a triangle $ABC$ intersect at $G$.

To prove that $ar(AGB)=ar(AGC)=ar(BGC)=\dfrac{1}{3} ar(ABC)$

Construction: Draw $BP \perp EG$

Proof: $A G=\dfrac{2}{3} A E \quad$ [Centroid divides the median in the ration 2:1]

Now, $ar(\triangle A G B)=\dfrac{1}{2} \times A G \times B P$

[Median divides a triangle into two triangles equal in area]

$=\dfrac{1}{3} ar(\triangle A B C)$

Again, $ar(\triangle A G C)=ar(\triangle B G C)=\dfrac{1}{3} ar(\triangle A B C)$

So, $ar(\triangle A G B)=ar(\triangle A G C)=ar(\triangle B G C)=\dfrac{1}{3} ar(\triangle A B C)$

Hence, proved.

Solution

Given: In triangle $A B C, X$ and $Y$ are the mid-points of $A B$ and $A C$.

To prove that $ar(ABP)=ar(ACQ)$.

Proof: Since, $X Y|| B Y$ [BY mid-point theorem]

As we know that triangle on the same base and between the same parallels lines are equal in area. So,

$ar(\triangle B Y C)=ar(\triangle B X C)$

Now, subtracting $ar(\triangle B O C)$ from both sides in the above, get:

$ar(\triangle B Y C)-ar(\triangle B O C)=ar(\triangle B X C)-ar(\triangle B O C)$

$ar(\triangle B O Y)=ar(\triangle C O X)$

Now, adding $ar(\triangle X O Y)$ to both side in equation (II), get:

$ar(\triangle B O Y)+ar(\triangle X O Y)=ar(\triangle C O X)+ar(\triangle X O Y)$

Again, quadrilaterals XYAP and YXAQ are on the same base XY and between the same parallels $X Y$ and $P Q$. So,

$ar(X Y A P)=ar(X Y Q A)$

Now, adding equation (III) and (IV), get:

$ar(\triangle B X Y)+ar(X Y A P)=ar(\triangle C X Y)+ar(X Y A Q)$

Hence, $ar(\triangle A B P)=ar(\triangle A C Q)$.

Solution

Given: $ABCD$ and $AEFD$ are two parallelogram.

To prove that $ar(\triangle P E A)=ar(\triangle Q F D)$

Construction: join PD.

Proof: In triangle PEA and triangle QFD,

$\angle A P E=\angle D Q F \quad$ [Corresponding angles are equal as $AB || CD]$

$\angle A E P=\angle D E Q \quad$ [Corresponding angles are equal as $AE || DF]$

$AE=DF \quad$ [Opposite sides of a parallelogram are equal]

So, $\triangle P E A \cong \triangle Q F D \quad$ [By AAS congruent rule]

Hence, $ar(\triangle P E A)=ar(\triangle Q E D)$.