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प्रश्न
ABC is a triangle in which ∠B = 2 ∠C. D is a point on BC such that AD bisects ∠BAC and AB = CD.
Prove that ∠BAC = 72°.
उत्तर
In ΔABC,
∠B = 2∠C or, ∠B = 2y, where ∠C = y.
AD is the bisector of ∠BAC.
So, let ∠BAD = ∠CAD = x.
Let BP be the bisector of ∠ABC. Join PD.
In ΔBPC, we have
∠CBP = ∠BCP = y ⇒ BP = PC
In Δ′s ABP and DCP, we have
∠ABP = ∠DCP,
∠ABP = ∠DCP = y
BP = PC ........(As proved above)
∠ADC = ∠ABD + ∠BAD
⇒ x + 2x = 2y + x
⇒ x = y
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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].
