+0  
 
0
721
1
avatar

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 $CM = \frac{1}{2} AB.$

 Jun 28, 2019
 #1
avatar+8965 
+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°

 Jun 28, 2019
edited by hectictar  Jul 15, 2019

14 Online Users

avatar
avatar