# Quadrilaterals

## Exercise 8.2

Question 1: ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal. Show that:

- SR||AC and `SR=1/2\AC`
- `PQ=SR`
- PQRS is a parallelogram

**Answer:** Let us extend the line SR to T so that CT is parallel to AS

In ΔDSR and ΔCRT

`DR=RC` (R is the midpoint of side DC)

`∠DRS=∠TRS` (opposite angles)

`∠DSR=∠RTC` (alternate angles of transversal ST when DA||CT)

Hence, `ΔDSR≅ΔCRT`

So, `SR=RT`

`ST=AC` (opposite sides of parallelogram)

So, `SR=1/2\AC`

As SR is touching the mid points of DA and DC so as per mid point theorem SR||AC

Similarly AC || PQ can be proven which will prove that PQRS is a parallelogram.

Question 2: ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

**Answer;** Following the method used in the previous question it can be proved that PQRS is a parallelogram. To prove it to be a rectangle we need to prove that

`∠S=∠R=∠Q=∠P=90°`

In ΔDSR, ΔCRQ, ΔBQP and ΔAPS

`DS=CR=BQ=AP=DR=CQ=BP=AS`

(All sides of rhombus are equal and PQRS are midpoints)

`∠DSR=∠DRS=∠CRQ=∠CQR=∠BQP=∠BPQ=∠APS=∠ASP`

So, `ΔDSR≅ ΔCRQ≅ ΔBQP≅ ΔAPS`

So, `∠SDR=∠CRQ=∠QBP=∠PAS=90°`

Hence, `∠DSR+∠DRS=90°`

Or, `∠DSR=∠DRS=∠CRQ=∠CQR=∠BQP=∠BPQ=∠APS=∠ASP`

As, `∠ASP+∠PSR+∠DSR=180°`

Or, `∠PSR=180°-(45°+45°)=90°`

Similarly, `∠S=∠R=∠Q=∠P=90°`

Hence, PQRS is a rectangle.

Question 3: ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F. Show that F is the mid-point of BC.

**Answer:** In Δ ADB

DG = GB

A parallel line to the base originating from mid point of second side will intersect at the midpoint of the third side.

AB || DC

AB || EF

So, EF || DC

So, In Δ ADB

EG || AB

E is the mid point of AD

So, G is the mid point of DB

Now, in Δ DCB

GF || DC

G is the mid point of BD

So, F will be mid point of BC ( Mid point theorem)