Solution

(i) In $\triangle A D C, S$ and $R$ are the mid-points of sides $A D$ and $C D$ 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.

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

(ii) In $\triangle A B C, P$ and $Q$ are mid-points of sides $A B$ and $B C$ respectively. Therefore, by using mid-point theorem,

$P Q || A C$ and $P Q=\frac{1}{2} A C \ldots$

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

$PQ || SR$ and $PQ=SR \ldots$ (3) $\angle$

$P Q=S R$ (iii) From equation (3), we obtained

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

Solution

b $\quad$ e

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

In $\triangle A D C$, In $\triangle A B C, P$ and $Q$ are the mid-points of sides $A B$ and $BC$ respectively.

AC (Using mid-point theorem) … (1)

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

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

$P Q || R S$ and $P Q=R S$

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 $A B C D$ intersect each other at point $O$.

In quadrilateral OMQN,

$\because$

$M Q || ON \because(PQ || AC) \quad QN$

I| OM ( QR || BD)

Therefore, OMQN is a parallelogram.

$\angle \angle \quad \angle M Q N=N O M$

$\angle L \quad \angle PQR=NOM$

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

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

Hence, $P Q R S$ is a rectangle.

Solution

Let us join $A C$ and $B D$.

In $\triangle ABC$,

$P$ and $Q$ are the mid-points of $A B$ and $B C$ respectively. $\angle PQ || AC$ and $\frac{1}{2} PQ=\quad AC($ Mid-point theorem $) \ldots$ Similarly in

$\triangle ADC$,

$\frac{1}{2}$

$SR || AC$ 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.

$\angle PS || QR$ and $PS=QR$ (Opposite sides of parallelogram) $\ldots(3)$

In $\triangle B C D, Q$ and $R$ are the mid-points of side $B C$ and $C D$ respectively.

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

However, the diagonals of a rectangle are equal. /

$A C=B D \ldots(5)$

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 $\triangle ABD$,

$E F || A B$ and $E$ is the mid-point of $A D$.

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

As $E F|| A B$ and $A B|| C D$,

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

In $\triangle BCD, GF || CD$ 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

$A B C D$ is a parallelogram.

$\angle A B | C D$

And hence, AE I| FC

Again, $A B=C D$ (Opposite sides of parallelogram ABCD)

$ _{A B} \frac{1}{2} CD $

$A E=F C(E$ and $F$ are mid-points of side $A B$ and $C D)$

In quadrilateral $A E C F$, one pair of opposite sides (AE and $C F$ ) is parallel and equal to each other. Therefore, AECF is a parallelogram. $\angle A F ||$ EC (Opposite sides of a parallelogram)

In $\triangle DQC, F$ 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.

$\angle DP=PQ \ldots(1)$

Similarly, in $\triangle A P B, E$ is the mid-point of side $A B$ and $E Q || A P$ (as $A F || E C$ ).

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

$\angle PQ=QB \ldots(2)$

From equations (1) and (2), DP

$=PQ=BQ$

Hence, the line segments $A F$ and $E C$ trisect the diagonal $B D$.

Solution

(i) In $\triangle A B C$,

It is given that $M$ is the mid-point of $A B$ and $M D || B C$.

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

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

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

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

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

$\angle M D A C$

(iii) Join MC.

In $\triangle AMD$ and $\triangle CMD$, $A D=C D$ ( $D$ is the mid-point of side $A C$ ) $\quad A D M=\angle C D M$ (Each 90)

$DM=DM$ (Common)

$\angle \triangle AMD \angle \triangle CMD$ (By SAS

Therefore, $A M=C M$ (By CPCT)

congruence rule)

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

Therefore, it fan be said that

$C M=A M={ }^{\frac{1}{2}} A B$