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