How to prove that angle bisector of right angle triangle ABC right angled at B is perpendicular bisector of third side AC.

Question

I have tried using sine theoram, angle bisector theoram, congruency of type RHS,AA,ASA but haven't been able to do this.

Answer 1

I don't think the statement you are trying to prove is true for any right triangle. Here is why.

Let the two adjacent sides of the right triangle be represented by vectors, A = (0,a), B = (b,0), then the hypotenuse of the triangle would be B-A = (b,-a) by the parallelogram law. Let the vector V = (1,1)(in fact, any vector with angle 45 degrees). Note that V is an angle bisector of the right angle.

Now, if we take the dot product of V and B-A, and set it equal to 0(since we want them to be perpendicular). We get b-a = 0. Thus V is perpendicular to the other side if and only if b = a.

Answer 2

Another proof that this is only true for a 45-45-90 right triangle.

If the angle bisector of the right angle is perpendicular to the hypotenuse, then the triangle formed by the bisector has a 45 angle (bisects the right angle) and a 90 angle (perpendicular) so the other angle has to be 45.

Therefore the original triangle is 45-45-90.