Solution

(i) In $\mathrm{△}ADC,S$ and $R$ are the mid-points of sides $AD$ and $CD$ respectively. In a triangle, the line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is half of it.

$\mathrm{\angle }SR||AC$ and $SR=\frac{1}{2}AC$.

(ii) In $\mathrm{△}ABC,P$ and $Q$ are mid-points of sides $AB$ and $BC$ respectively. Therefore, by using mid-point theorem,

$PQ||AC$ and $PQ=\frac{1}{2}AC\dots$

Using equations (1) and (2), we obtain

$PQ||SR$ and $PQ=SR\dots$ (3) $\mathrm{\angle }$

$PQ=SR$ (iii) From equation (3), we obtained

Clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal.

Solution

b $$ e

$\mathrm{\angle }PQ||AC$ and $PQ=\stackrel{\frac{1}{2}<}{2}$

In $\mathrm{△}ADC$, In $\mathrm{△}ABC,P$ and $Q$ are the mid-points of sides $AB$ and $BC$ respectively.

AC (Using mid-point theorem) … (1)

$R$ and $S$ are the mid-points of $CD$ and $AD$ respectively. $RS||AC$ and $RS=\sqrt{\frac{1}{2}}AC$ (Using mid-point theorem)

From equations (1) and (2), we obtain

$PQ||RS$ and $PQ=RS$

Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a parallelogram.

Let the diagonals of rhombus $ABCD$ intersect each other at point $O$.

In quadrilateral OMQN,

$\because$

$MQ||ON\because (PQ||AC)QN$

I| OM ( QR || BD)

Therefore, OMQN is a parallelogram.

$\mathrm{\angle }\mathrm{\angle }\mathrm{\angle }MQN=NOM$

$\mathrm{\angle }L\mathrm{\angle }PQR=NOM$

However, $\mathrm{\angle }NOM={90}^{\circ }$ (Diagonals of a rhombus are perpendicular to each other) $\mathrm{\angle }$ $\mathrm{\angle }PQR={90}^{\circ }$

Clearly, PQRS is a parallelogram having one of its interior angles as ${90}^{\circ }$.

Hence, $PQRS$ is a rectangle.

Solution

Let us join $AC$ and $BD$.

In $\mathrm{△}ABC$,

$P$ and $Q$ are the mid-points of $AB$ and $BC$ respectively. $\mathrm{\angle }PQ||AC$ and $\frac{1}{2}PQ=AC($ Mid-point theorem $)\dots$ Similarly in

$\mathrm{△}ADC$,

$\frac{1}{2}$

$SR||AC$ and $SR=AC$ (Mid-point theorem) … (2)

Clearly, $PQ||SR$ and $PQ=SR$

Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a parallelogram.

$\mathrm{\angle }PS||QR$ and $PS=QR$ (Opposite sides of parallelogram) $\dots (3)$

In $\mathrm{△}BCD,Q$ and $R$ are the mid-points of side $BC$ and $CD$ respectively.

$\mathrm{\angle }QR||BD$ and $QR=\frac{1}{2}BD$ (Mid-point theorem) …

However, the diagonals of a rectangle are equal. /

$AC=BD\dots (5)$

By using equation (1), (2), (3), (4), and (5), we obtain PQ

$=QR=SR=PS$ Therefore, $PQRS$ is a rhombus.

Solution

Let $EF$ intersect $DB$ at $G$.

By converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle and parallel to another side, bisects the third side.

In $\mathrm{△}ABD$,

$EF||AB$ and $E$ is the mid-point of $AD$.

Therefore, $G$ will be the mid-point of $DB$.

As $EF||AB$ and $AB||CD$,

$\mathrm{\angle }EF||CD$ (Two lines parallel to the same line are parallel to each other)

In $\mathrm{△}BCD,GF||CD$ and $G$ is the mid-point of line $BD$. Therefore, by using converse of mid-point theorem, $F$ is the mid-point of $BC$.

Solution

$ABCD$ is a parallelogram.

$\mathrm{\angle }AB|CD$

And hence, AE I| FC

Again, $AB=CD$ (Opposite sides of parallelogram ABCD)

${}_{AB}\frac{1}{2}CD$

$AE=FC(E$ and $F$ are mid-points of side $AB$ and $CD)$

In quadrilateral $AECF$, one pair of opposite sides (AE and $CF$ ) is parallel and equal to each other. Therefore, AECF is a parallelogram. $\mathrm{\angle }AF||$ EC (Opposite sides of a parallelogram)

In $\mathrm{△}DQC,F$ is the mid-point of side $DC$ and $FP||CQ$ (as $AF||EC$ ). Therefore, by using the converse of mid-point theorem, it can be said that $P$ is the mid-point of

DQ.

$\mathrm{\angle }DP=PQ\dots (1)$

Similarly, in $\mathrm{△}APB,E$ is the mid-point of side $AB$ and $EQ||AP$ (as $AF||EC$ ).

Therefore, by using the converse of mid-point theorem, it can be said that $Q$ is the mid-point of $PB$.

$\mathrm{\angle }PQ=QB\dots (2)$

From equations (1) and (2), DP

$=PQ=BQ$

Hence, the line segments $AF$ and $EC$ trisect the diagonal $BD$.

Solution

(i) In $\mathrm{△}ABC$,

It is given that $M$ is the mid-point of $AB$ and $MD||BC$.

Therefore, $D$ is the mid-point of $AC$. (Converse of mid-point theorem)

(ii) As DM || $CB$ and $AC$ is a transversal line for them, therefore,

$\mathrm{\angle }MDC+\mathrm{\angle }DCB={180}^{\circ }$ (Co-interior angles)

$\mathrm{\angle }MDC+{90}^{\circ }={180}^{\circ }$

$\mathrm{\angle }MDC={90}^{\circ }$

$\mathrm{\angle }MDAC$

(iii) Join MC.

In $\mathrm{△}AMD$ and $\mathrm{△}CMD$, $AD=CD$ ( $D$ is the mid-point of side $AC$ ) $ADM=\mathrm{\angle }CDM$ (Each 90)

$DM=DM$ (Common)

$\mathrm{\angle }\mathrm{△}AMD\mathrm{\angle }\mathrm{△}CMD$ (By SAS

Therefore, $AM=CM$ (By CPCT)

congruence rule)

However, $AM=\stackrel{\frac{1}{2}}{AB}$ (M is the mid-point of $.AB{)}^{\mathrm{\angle }}$

Therefore, it fan be said that

$CM=AM={}^{\frac{1}{2}}AB$