Question

The line segment joining the mid – points M and N of parallel

sides AB and DC respectively of a trapezium ABCD is

perpendicular to both the sides AB and DC. Prove that AD = BC.​

Answer 1

Answer:

Construct AN and BN at the point N

Consider △ANM and ∠BNM

We know that N is the midpoint of the line AB

So we get

AM=BM

From the figure we know that

∠AMN=∠BMN=90

∘

MN is common i.e. MN=MN

By SAS congruence criterion

△ANM≅△BNM

AN=BN(c.p.c.t)…(1)

We know that

∠ANM=∠BNM(c.p.c.t)

Subtracting LHS and RHS by 90

∘

90

∘

−∠ANM=90

∘

−∠BNM

So we get

∠AND=∠BNC…(2)

Now, consider △AND and △BNC

AN=BN

∠AND=∠BNC

We know that N is the midpoint of the line DC

DN=CN

By SAS congruence criterion

△AND≅△BNC

AD=BC(c.p.c.t)

Therefore, it is proved that AD=BC