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:

quadrilaterals 2

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.

quadrilaterals 4

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.

quadrilaterals 6

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.

quadrilaterals 7
quadrilaterals 8

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.

quadrilaterals 9

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

quadrilaterals 11

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