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ΔABC is isosceles with AB = AC. Also. AD ⊥ BC meeting BC in D. Are the two triangles ABD and ACD congruent? State in symbolic form. Which congruence condition do you use? Which side of ADC equals BD? Which angle of Δ ADC equals ∠B?

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We have AB = AC …… (i)

AD = DA (common) …… (ii)

And, ∠ADC = ∠ADB (AD ⊥ BC at point D) …… (iii)

Therefore, from (i), (ii) and (iii), by RHS congruence condition, ΔABD ≅ ΔACD, the triangles are congruent.

Therefore, BD = CD.

And ∠ABD = ∠ACD (corresponding parts of congruent triangles)

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