We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
82
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+8406 
+2

(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

7 Online Users