SOLUTION: In △ABC, m∠CAB = 60� and point D∈BC so that AD=10 in and the distance from D to AB is 5 in. Prove that AD is the angle bisector of ∠A.

Question 1127226: In △ABC, m∠CAB = 60� and point D∈BC so that AD=10 in and the distance from D to AB is 5 in. Prove that AD is the angle bisector of ∠A.
Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!

If AD is 10 and the distance from D to AB is 5, then triangle ABD is a 30-60-90 right triangle, with the 30-degree angle at A.

(My sketch has AB horizontal, with B to the right of A; AD slanting up to the right from A; DB perpendicular to AB.)

Since angle A is 60 degrees, AD bisects it.
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SOLUTION: In &#9651;ABC, m&#8736;CAB = 60� and point D&#8712;BC so that AD=10 in and the distance from D to AB is 5 in. Prove that AD is the angle bisector of &#8736;A.

SOLUTION: In △ABC, m∠CAB = 60� and point D∈BC so that AD=10 in and the distance from D to AB is 5 in. Prove that AD is the angle bisector of ∠A.