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Question
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°.
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Solution
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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