Question

In a right triangle, prove that the line-segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.

Answer 1

Given: In ΔABC, ∠B = 90° and D is the mid-point of AC.

Construction: Produce BD to E such that BD = DE and join EC.

To prove: BD = `1/2` AC

Proof: In ΔADB and ΔCDE,

AD = DC   ...[∵ D is mid-point of AC]

BD = DE   ...[By construction]

And ∠ADB = ∠CDE   ...[Vertically opposite angles]

∴ ΔADB ≅ ΔCDE   ...[By SAS congruence rule]

⇒ AB = EC   ...[By CPCT]

And ∠BAD = ∠DCE   ...[By CPCT]

But ∠BAD and ∠DCE are alternate angles.

So, EC || AB and BC is a transversal.

∴ ∠ABC + ∠BCE = 180°  ...[Cointerior angles]

⇒ 90° + ∠BCE = 180°   ...[∵ ∠ABC = 90°, given]

⇒ ∠BCE = 180° – 90°

⇒ ∠BCE = 90°

In ΔABC and ΔECB,

AB = EC  ...[Proved above]

BC = CB   ...[Common side]

And ∠ABC = ∠ECB   ...[Each 90°]

∴ ΔABC ≅ ΔECB   ...[By SAS congruence rule]

⇒ AC = EB   ...[By CPCT]

⇒ `1/2` EB = `1/2` AC   ...[Dividing both sides by 2]

⇒ BD = `1/2` AC   

Hence proved.