SOLUTION: 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.

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Question 1102480: 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.
Answer by ikleyn(52818)   (Show Source): You can put this solution on YOUR website!
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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.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

0.  Make a sketch to follow my proof.


1.  The fact that "the distance from D to AB is 5 in" means 

    that the length of the perpendicular drawn from the point D to the side AB is 5 inches.


    Draw this perpendicular. Let E be the intersection point of this perpendicular with the side AB.


    You have a right-angled triangle ADE.  Its hypotenuse AD is 10 in long.

    Its leg DE is 5 inches long.

    Hence, the angle EAD is 30°.

    The measure of the angle EAD is the same as the measure of the angle BAD.

    It implies that AD is the angle bisector of the angle BAC.

QED.


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