Question

Prove the Following! Need help ASAP!

Answer 1

Since AD is the angle bisector of ∠A and ΔABC is an isosceles triangle:

=> AD ⊥ BC

=> ∠ADB = ∠ADC = 90°

We have: ∠BAD = ∠CAD (AD is an angle bisector of ∠A)

∠ADB = ∠ADC = 90°

Both triangles have AD in common

=> ΔABD ≅ ΔACD

Answer 2

Given:

ΔABC is an isosceles triangle

AD is an angle bisector of ∠A

Statement | Proof

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ΔABC is an isosceles triangle Given; Definition of an

isosceles triangle.

AD bisects ∠ABC Definition of a angle bisector.

AD ≅ AD Reflexive Property

AD bisects BC Definition of a line segment bisector

BD ≅ CD " "

∠BDA ≅ ∠ADC Triangle Bisector Theorem

What has been proven so far:

AD ≅ AD ; ∠BDA ≅ ∠CDA ; BD ≅ CD

Based on this information,

ΔABD ≅ ΔACD Side-Angle-Side Theorem

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