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)

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.

**Answer:** In ΔADE and ΔCBF

AD = BC (Opposite sides of parallelogram)

BF = DE (Half of opposite sides of parallelogram)

∠ADE = ∠CBF (Opposite angles)

So, ΔADE ≅ ΔCBF

Hence, AE = CF

In quadrilateral AECF

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`

Copyright © excellup 2014