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

Jul 7, 2019

#1
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(b)

Let     AB  =  c     (because it is the side across from angle C)

and    AC  =  b     (because it is the side across from angle B)

and    BC  =  a     (because it is the side across from angle A)

Draw a height from  M  which meets side  AC  at point  D

m∠BAC  =  m∠MAD     because they are the same angle

m∠ACB  =  m∠ADM     because they are both right angles

So by AA similarity,  △ABC ~ △AMD

And we know   AM  =  c / 2   because  M  is the midpoint of  AB

So the scale factor from  △ABC  to  △AMD  is  1/2   And so...

AM  =  c / 2

DM  =  a / 2

Then by SAS congruence we can determine that  △ADM  ≅  △CDM   and so...

CM  =  AM

CM  =  c / 2

CM  =  (1/2)(AB)

Jul 7, 2019

#1
+8966
+5

(b)

Let     AB  =  c     (because it is the side across from angle C)

and    AC  =  b     (because it is the side across from angle B)

and    BC  =  a     (because it is the side across from angle A)

Draw a height from  M  which meets side  AC  at point  D

m∠BAC  =  m∠MAD     because they are the same angle

m∠ACB  =  m∠ADM     because they are both right angles

So by AA similarity,  △ABC ~ △AMD

And we know   AM  =  c / 2   because  M  is the midpoint of  AB

So the scale factor from  △ABC  to  △AMD  is  1/2   And so...

AM  =  c / 2

DM  =  a / 2

Then by SAS congruence we can determine that  △ADM  ≅  △CDM   and so...

CM  =  AM

CM  =  c / 2

CM  =  (1/2)(AB)

hectictar Jul 7, 2019