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 Jul 7, 2019

#1**+5 **

**(a)** https://web2.0calc.com/questions/math-halp-plz#r1

**(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

AD = b / 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

#1**+5 **

Best Answer

**(a)** https://web2.0calc.com/questions/math-halp-plz#r1

**(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

AD = b / 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