Question

3. Prove that the quadrilateral obtained by joining the mid-points of an isosceles
trapezium is a rhombus.
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Answer 1

Answer:

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Step-by-step explanation:

Given: ABCD is an isosceles trapezium in which AB∣∣DC and AD=BC

P,Q,R and S are the mid-points of the sides AB,BC,CD and DA respectively PQ,QR,RS and SP are joined.

To prove: PQRS is a rhombus

Construction: Join AC and BD

 Proof:

Since, ABCD is an isosceles trapezium

Its diagonals are equal

AC=BD

Now, in △ABC

P and Q are the mid-points of AB and BC

So, PQ∣∣AC and

PQ=21AC … (i)

Similarly, in △ADC

S and R mid-point of CD and AD

So, SR∣∣AC and SR=21AC … (ii)

From (i) and (ii), we have

PQ∣∣SR and PQ=SR

Thus, PQRS is a parallelogram.

Now, in △APS and △BPQ

AP=BP [P is the mid-point]

AS=BQ [Half of equal sides]

∠∠A=∠∠B [As ABCD is an isosceles trapezium]

So, △APS≅△BPQ by SAS Axiom of congruency

Thus, by C.P.C.T we have

PS=PQ

But there are the adjacent sides of a parallelogram

So, sides of PQRS are equal

Hence, PQRS is a rhombus

Hence proved