Question

In a quadrilateral ABCD , M & N are mid point of AB & CD , MN is perpendicular on both AB & CD Prove that AD = BC​

Answer 1

Given
ABCD is a trapezium
M is the mid point of AB
N is the mid point of CD

Construction
Join B to N
Join A to N
Proof
Consider ΔAMN and ΔBMN
∠AMN=∠BMN=90
AM=BM (M is the midpoint of AB)
MN=MN(common)
ΔAMN congruent to ΔBMN(SAS congruence rule)

Consider ΔADN and ΔBCN
DN=CN(N is the midpoint of CD)
AN=BN(CPCT)
∠MNA=∠BNM(CPCT) 1
∠MNC=∠MND= 90 2
2-1
∠MND-∠MNA=∠MNC-∠BNM
∠AND=∠BNC
ΔAND congruent to ΔBNC
AD=BC(CPCT)
Hence proved

Hope its helpful.....