# Thread: An analytical geometry question!

1. ## An analytical geometry question!

Hey guys! So I have this question in my worksheet and for some reason it sounds completely foreign to me. I think my teacher overlooked this kind of question, so I'm pretty stuck...

"Given right triangle ABC with D the midpoint of BA.. Show that D is equidistant from B and C."

and here is a "lovely" illustration to go with it!

So yeah, any help would be greatly appreciated. Thanks so much!

2. Originally Posted by Shnub
Hey guys! So I have this question in my worksheet and for some reason it sounds completely foreign to me. I think my teacher overlooked this kind of question, so I'm pretty stuck...

"Given right triangle ABC with D the midpoint of BA.. Show that D is equidistant from B and C."

and here is a "lovely" illustration to go with it!

So yeah, any help would be greatly appreciated. Thanks so much!

Hello , Shnub

We first confirm the coordinates of point $D$ , by using mid-pt formula we have $D = D\left( \frac{a+0}{2} , \frac{0+b}{2} \right) = D\left( \frac{a}{2} , \frac{b}{2} \right)$

Then , use the distance formula , the distance from $D$ to the right angle $C$is :

$\sqrt{ (\frac{a}{2} - 0)^2 + ( \frac{b}{2} - 0)^2 } = \sqrt{( \frac{a}{2})^2 + (\frac{b}{2})^2 }$

and the distance $AD$ is $\sqrt{ (a - \frac{a}{2} )^2 + (0 - \frac{b}{2})^2 } = \sqrt{ ( \frac{a}{2})^2 + (\frac{b}{2})^2 }$

Can you see something special ?

3. Hello, Shnub!

Given right triangle $ABC$ with $D$ the midpoint of hypotenuse $AB$,
show that $D$ is equidistant from $B$ and $C.$
Code:
              * * *
*           *   C
*               o
*              *  *
*
*           *       *
A o - - - - o - - - - o B
*         D         *

*                 *
*               *
*           *
* * *

Draw $AC$ and $BC.$

A right triangle can be inscribed in a semicircle.

The midpoint of the hypotenuse is the center of the circle.

Therefore: . $DA = DB = DC = \text{radius}$