# Why is AM=MB=MC in this right triangle?

## I understand that AM=BC because M is the centerpoint, but I do not understand MC.

Jul 20, 2018

#### Explanation:

$M$ is the midpoint of $A B$ and $\angle A C B = {90}^{\circ}$.

Therefore, $C$ lies on the (semi)circle described on $A B ,$ taking $A B$

as diameter.

$M$ is the midpoint of the diameter $A B$.

$\therefore M$ is the centre of the circle.

Thus, $C$ lies on the circle having centre at $M$ and a diameter $A B$.

Clearly, $M A = M B = M C \text{(=the radius of the circle)}$.

Jul 20, 2018

#### Explanation:

I understand that $A M = \textcolor{red}{B C}$ because $M$ is

the centerpoint, but I do not understand $M C .$

I think it is $A M = M B \mathmr{and}$ not $\textcolor{red}{B C}$

We have two Theorem :

color(green)((1) "An angle inscribed in a semicircle is a right angle"

color(green)((2)"The circle whose diameter is the hypotenuse of the "

color(green)("right triangle , passes through three vertices of the triangle."

So, $A , B , \mathmr{and} C$ are the points on the circle and $m \angle C = {90}^{\circ}$

:.color(violet)( " Hypotenuse AB of triangle = Diameter AB of circle"

Now , $M \text{ is the midpoint of AB}$

So, $\text{M is the center of the circle. }$

Let $\textcolor{b l u e}{r}$ be the radius of circle.

:.color(blue)(AM= MB=r

Here, $\overline{M C}$ is the line segment joining center $M$ and vertex $C$

$\therefore M C \text{ is the radius of circle.}$

$\therefore \textcolor{b l u e}{M C = r}$

Hence ,

color(blue)(AM=MB=MC=r