Question

ABC is an isosceles triangle with AB = AC and D is a point on BC such that AD ⊥ BC (Figure). To prove that ∠BAD = ∠CAD, a student proceeded as follows:

In ∆ABD and ∆ACD,

AB = AC (Given)

∠B = ∠C (Because AB = AC)

and ∠ADB = ∠ADC

Therefore, ∆ABD ≅ ∆ACD (AAS)

So, ∠BAD = ∠CAD (CPCT)

What is the defect in the above arguments?

[Hint: Recall how ∠B = ∠C is proved when AB = AC].

Answer 1

In ∆ABC, AB = AC

⇒ ∠ACB = ∠ABC   ...[Angles opposite to the equal sides are equal]

In ∆ABD and ∆ACD,

AB = AC   ...[Given]

∠ABD = ∠ACD   ...[Proved above]

∠ADB = ∠ADC   ...[Each 90°]

∴ ∆ABD ≅ ∆ACD   ...[By AAS]

So, ∠BAD = ∠CAD   ...[By CPCT]

So, the defect in the given argument is that firstly prove ∠ABD = ∠ACD

Hence, ∠ABD = ∠ACD is defect.