Question

The line segments joining the midpoints 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

Given: In trapezium ABCD, M and N are mid-points of AB and DC, MN ⊥ AB and MN ⊥ DC .To prove: AD = BC
Construction: Join CM and DM.
Proof:
In Δ CMN and Δ DMN ,
MN = MN
(Common sides)
∠ CNM = ∠ DNM = 90° (Given, MN ⊥ DC )
CN = DN
(Given, N is the mid-point DC )
∴ By SAS congruence criteria,
Δ CMN ≅ Δ DMN
So, CM = DM
(CPCT) .....(i)
And, ∠ CMN = ∠ DMN (CPCT)
But, ∠ AMN = ∠ BMN = 90°
(Given, MN ⊥ AB)
⇒∠ AMN - ∠ CMN = ∠ BMN - ∠ DMN
⇒∠ AMD = ∠ BMC
.....(ii)
Now, in Δ AMD and Δ BMC,
DM = CM
∠ AMD = ∠ BMC
AM = BM
[From (i)]
[From (ii)]
(Given, M is the mid-point AB)
∴ By SAS congruence criteria,
Δ AMD ≅ Δ BMC
Hence, AD = BC (CPCT)