Question

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

Given: In trapezium ABCD, points M and N are the mid-points of parallel sides AB and DC respectively and join MN, which is perpendicular to AB and DC.

To prove: AD = BC

Proof: Since, M is the mid-point of AB.

∴ AM = MB

Now, in ΔAMN and ΔBMN,

AM = MB   ...[Proved above]

∠3 = ∠4   ...[Each 90°]

MN = MN  ...[Common side]

∴ ΔAMN ≅ ΔBMN   ...[By SAS congruence rule]

∴ ∠1 = ∠2    ...[By CPCT]

On multiplying both sides of above equation by –1 and then adding 90° both sides, we get

90° – ∠1 = 90° – ∠2

 ⇒ ∠AND = ∠BNC  ...(i)

Now, in ΔADN and ΔBCN,

∠AND = ∠BNC   ...[From equation (i)]

AN = BN   ...[∵ΔAMN ≅ ΔBMN]

And DN = NC   ...[∵ N is the mid-point of CD (given)]

∴ ΔADN ≅ ΔBCN   ...[By SAS congruence rule]

Hence, AD = BC   ...[By CPCT]

Hence proved.