In triangle $ABC,$ let $M$ be the midpoint of $\overline{AB}.$ (a) Prove that if $CM = \frac{1}{2} AB,$ then $\angle ACB = 90^\circ.$ (b) Prove that if $\angle ACB = 90^\circ,$ then $CM = \frac{1}{2} AB.$

Guest Jun 28, 2019

#1**+5 **

**(a)**

Let the length of side AB = c

and let m∠ABC = B

and let m∠BAC = A

We're given that \(CM = \frac{1}{2} AB\) so

CM = c / 2

And because M is the midpoint of side AB,

AM = c / 2

BM = c / 2

Therefore both △AMC and △BMC are isosceles triangles. Because base angles are congruent,

m∠MCA = A

m∠MCB = B

And now we can see that

m∠ACB = A + B

The sum of the interior angles of △ABC = 180°

(A) + (A + B) + (B) = 180°

2A + 2B = 180°

2(A + B) = 180°

A + B = 90°

m∠ACB = 90°

hectictar Jun 28, 2019