Question

Question #f7b8b

Answer 1

None of the four answers suggested in the problem is applicable.

Explanation:

The angle bisector theorem deals with an angle bisector in a triangle. The theorem states that this angle bisector divides the opposite to angle side proportionally to corresponding adjacent sides.
This is not what our problem is about.

Converse to this theorem states that if we divide a side of a triangle proportionally to two other sides than a segment from the opposite angle to a point of division bisects the angle.
Also not related to our problem.

Perpendicular bisector theorem and its converse deal with a segment divided in halves by a perpendicular line.
Does not seem related to our problem at all.

You might be interested in the proof of a theorem stated in the problem. It's easy.
Consider an angle `/_BAC` and point `D` on its bisector.

Assume that `DB` is perpendicular to `AB` and `DC` is perpendicular to `AC`.
Let's prove that `DB` and `DC` are congruent (that is, point `D` is equidistant from sides of an angle `/_BAC`

PROOF
Consider triangles `Delta ADB` and `Delta ADC`.
Both are right triangles, `/_ABD = /_ACD = 90^o`.
They have congruent acute angles `/_BAD` and `/_CAD` (since `AD` is an angle bisector).
They share hypotenuse `AD`.
Therefore, these right triangles are congruent by two angles and a side and, therefore, their sides `DB` and `DC` are congruent too, as lying across congruent angles.
End of proof.