Question

Prove the Following! Need help ASAP!

Answer 1

Answer:

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

Answer:

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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