Prove that, if the bisector of $∠$ BAC of $∆$ ABC is perpendicular to side BC, then $∆$ ABC is an isosceles triangle.

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Solution

In △ADB and △ADC
∠BAD = ∠CAD (AD is the bisector of ∠BAC)
∠BDA = ∠CDA = 90^∘ (Given)
AD = DA (Common)
By ASA test of congruency
△ADB ≅ △ADC
∴ ∠B = ∠C (corresponding angles of congruent triangles)
If two angles of a traingle are equal, then the traingle is said to be a isosceles triangle.

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Prove that, if the bisector of ∠BAC of ∆ABC is perpendicular to side BC, then ∆ABC is an isosceles triangle.

Prove that, if the bisector of $∠$ BAC of $∆$ ABC is perpendicular to side BC, then $∆$ ABC is an isosceles triangle.