## 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.

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)

Question 4: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively. Show that the line segments AF and EC trisect the diagonal BD.

AD = BC (Opposite sides of parallelogram)

BF = DE (Half of opposite sides of parallelogram)

∠ADE = ∠CBF (Opposite angles)

So, ΔADE ≅ ΔCBF

Hence, AE = CF
EC || AF & EC = AF
AE = CF
So, AE || CF
So, AECF is a parallelogram.
In Δ DQC
PE || QC (proved earlier by proving AE || CF)
E is the mid point of DC
So, P is the mid point of DQ
So, DP = PQ
In Δ APB
FQ || AP
F is the mid point of AB
So, PQ = QB
So, DP = PQ = QB proved

Question 5: Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Answer: ABCD is a quadrilateral in which P, Q, R, & S are mid points of AB, BC, CD & AD
In Δ ACD
SR is touching mid points of CD and AD
So, SR || AC
Similarly following can be proved
PQ || AC
QR || BD
PS || BD
So, PQRS is a parallelogram.
PR and QS are diagonals of the parallelogram PQRS, so they will bisect each other.

Question 6: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that

• D is the midpoint of AC
• MD ⊥ AC
• CM=MA=1/2\AB

Answer: DM || BC
M is the mid point of AB
So, D is the mid point of AC (Mid point theorem)

∠ACD=∠MDA=90° (alternate angle to transversal MD)
Now in ΔCDM and ΔADM
CD=AD
MD=MD
∠MDC=∠MDA
So, ΔCDM≅ ΔADM (SAS theorem)
So, MC=MA
MA=1/2\AB
So, MC=MA=1/2\AB