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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.
Solution
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.
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