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 $CM = \frac{1}{2} AB.$
+9465 (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°
