# Triangle inscribed with an angle of 60°

• November 20th 2008, 11:12 AM
lego_lax
Triangle inscribed with an angle of 60°
I have $\Delta ABC$ inscribed in a circle with $\angle A \:=\:60^o$. The height relative to BC is equal to the radius. On the line BA take the segment AD equal to AC. I have to prove that the line DC is tangent to the circle.
• November 22nd 2008, 12:45 PM
Soroban
Hello, lego_lax!

If no one is responding, it's that we didn't understand the description.

After a number of incorrect sketches, I finally caught on.

Quote:

I have $\Delta ABC$ inscribed in a circle with $\angle A \:=\:60^o$.
The height relative to BC is equal to the radius.
On the line BA take the segment AD equal to AC.
Prove that the line DC is tangent to the circle.

Code:

```                      D                       *                     /                     /                   /                   /                 A/               * o *           *    / \*  *         *    /  \ *  *       *    /    \  *  *             /      r\  * *       *    /        \    o C       *  /          o  *       *  /      *    E  *         /  *     B o                *         *              *           *          *               * * *```

We have inscribed triangle $ABC$, with $\angle BAC = 60^o.$

$AE \perp BC,\;\;AE = r.$

Extend $BA$ to $D$ so that $AD = AC.$

Prove that $DC$ is tangent to the circle.

Now I can give it a try . . .

• November 22nd 2008, 01:44 PM
lego_lax
Quote:

Originally Posted by Soroban
Hello, lego_lax!

If no one is responding, it's that we didn't understand the description.

Thanks.
I'm sorry but my English is not so good.
However, I have already solved the problem but is a bit too complicated to write for me.
Prove that the side AB must necessarily go through the center of the circle.