Q6 of 49 Page 47

Prove that a ΔABC is an isosceles triangle if the altitude AD from A on BC bisects BC.

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Δ ABC is isosceles with AB = AC


AD is the altitude on BC


In Δ ABD and Δ ACD


∠ABD = ∠ACD (Base angles of an isosceles triangle)


∠ADB = ∠ADC (AD is a perpendicular)


AD = AD (Common side)


So Δ ABD and Δ ACD are congruent to each other by A.A.S. axiom of congruency


BD = DC(Corresponding Parts of Congruent Triangles)


Since BD = DC so D is the midpoint of BC


So altitude AD bisects BC


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