# hypotenuse of a right triangle

• Mar 25th 2007, 09:33 AM
JayJay1206
hypotenuse of a right triangle
Hi,

I need help with this problem. May someone help me please?

The Problem is: You are asked to prove the theorem given below, using the figure shown. What specifically do you need to prove in the figure?

The median drawn to the hypotenuse of a right triangle divides it into two isosceles triangles.

• Mar 25th 2007, 10:28 AM
ThePerfectHacker
Thou art asked to show that,
triangle BAD and triangle CAD are isoseles triangles.

Use the following theorem.
Theorem: Given a triangle ABC. From A draw a median to BC. Let is intersect at D.
Then,

Now the triangle which we have is right.
Hence,
AB^2+AC^2 = BC^2 by Pythagorean Theorem.

Thus,
But (1/2)BC=BD=DC
Hence,
Hence the two triangles are isoseles.
Q.E.D.
• Mar 25th 2007, 10:43 AM
Soroban
Hello, JayJay!

Quote:

Prove: the median drawn to the hypotenuse of a right triangle
divides it into two isosceles triangles.

There is an "eyeball" proof for this.
Maybe you can explain it in words.

Code:

```                * * *      C           *              o         *            o    o*       *          o        o*               o              o       *  o                  *     A o  o  o  *  o  o  o B       *    r    O    r    *       *                    *         *                  *           *              *                 * * *```

A right triangle can be inscribed in a semicircle.

We have right triangle ABC inscribed in a semicircle with radius r.
. . Hence: OA = OB = r.

Draw radius OC and we have:
. . OC = OA . . ∆AOC is isosceles.
. . OC = OB . . ∆COB is isosceles.

• Mar 25th 2007, 06:54 PM
JayJay1206
Thank you