Question

In a right-angles triangle ABC, D is the midpoint of the hypotenuse AC. Prove that BD-AC.​

Answer 1

Let ∆ABC is a right triangle such that ∠B = 900 and D is mid point of AC then we have to prove that BD = 1/2 AC we produce BD to E such that BD = AC and EC.

Now in ∆ADB and ∆CDE we have

AD = DC [Given]

BD = DE [By construction]

And, ∠ADB = ∠CDE [Vertically opposite angles] ∴ By SAS criterion of congruence we have

∆ADB ≅ ∆CDE

EC = AB and ∠CED = ∠ABD ....(i) [By cpctc]

But ∠CED & ∠ABD are alternate interior angles

∴ CE ║ AB

⇒ ∠ABC + ∠ECB = 1800 [Consecutive interior angles]

⇒ 90 + ∠ECB = 1800

⇒ ∠ECB = 900

Now, In ∆ABC & ∆ECB we have

AB = EC [By (i)]

BC = BC [Common]

And, ∠ABC = ∠ECB = 900

∴ BY SAS criterion of congruence

∆ABC ≅ ∆ECB

⇒ AC = EB [By cpctc]

⇒ 1/2 AC = 1/2 EB

⇒ BD = 1/2 AC

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