Thread: 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.

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,
AD = (1/2)*sqrt(2(AB^2+AC^2)-BC^2)

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

Thus,
AD=(1/2)*sqrt(2BC^2-BC^2)=(1/2)*sqrt(BC^2)=(1/2)BC
But (1/2)BC=BD=DC
Hence,
AD=BD and AD=DC
Hence the two triangles are isoseles.
Q.E.D.

Hello, JayJay!

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.

Thank you