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