Question

prove that line segment joining the midpoint of the parallel sides of a trapezium divides it in two equal parts

Answer 1

Answer:

Step-by-step explanation:

Given: PQRS is a trapezium and M,N are the midpoints on PQ an SR respectively.

To prove: Line segment joining the midpoint of the parallel sides of a trapezium divides it in two equal parts.

Construction: PA and PB are heights of the trapezium.

Proof: Let us assume that PQRS is an isosceles triangle with PS=QR.

Now, PM=MQ, SN=NR

and AN=PM, MQ=NB( opposite angles of the rectangle)

Therefore, ar(PMNA)=ar(MQBN)

Now,in ΔPSA and ΔQRB,

PA=QB(Heights of trapezium)

PS=QR (assumption)

∠PAS=∠QBR ( each 90°)

By RHS rule,

ΔPSA ≅ ΔQRB

⇒ar(PAS)=ar(QBR)

Also, ar(PMNS)=ar(PAS)+ar(PMNA)

and ar(MQRN)=ar(QBR)+ar(MQBN)

⇒ar(PMNS)=ar(MQRN)

Hence proved.