Question 1102352: In △ABC, m∠CAB = 60° and point D ∈ BC so that AD = 10 in, and the distance from D to AB is 5in. Prove that AD is the angle bisector of ∠A. Which theorem used to solve this problem?
a) Leg opposite to 30°∠
b) Leg opposite to 30°∠ converse
c) Median to hypotenuse theorem
d) Median to hypotenuse theorem converse
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
The names of the theorems you show as the answer choices are not names that are used universally, or probably even widely. There is no median to a hypotenuse in this problem, so answer choices c and d are out. But whether the answer is a or b depends on how the theorems were presented to you in your book and/or by your teacher.
It is clear that AD bisects angle A; what the answer is to the problem only you can figure out.
It is given that angle A is a 60 degree angle.
We also know that AD is 10 and BD is 5; also, since BD is the distance from D to AB, we know BD is perpendicular to AB. That makes triangle ABD a 30-60-90 right triangle, with angle DAB 30 degrees.
Then since angle BAD is 30 degrees and angle A is 60 degrees, AD bisects angle A.
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