Question

show that quadrilateral formed by joining midpoints of isosceles trapezium is rhombus. pls help me fast!

Answer 1

Construct the diagonals in the trapezuim like AC and BD. By midpoint theorem the opposite sides of the quad. obtained by joining the midpoints will come equal to each other and half of the diagonal in between the opposite sides. Then by congruency the two diagonals will come equal and all the sides will come equal.
Hence proved it is a rhombus.

Answer 2

Answer:

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=

2

1

AC … (i)

Similarly, in △ADC

S and R mid-point of CD and AD

So, SR∣∣AC and SR=

2

1

AC … (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

solution