Class 9 Maths

## 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.