# tangents

• Jul 8th 2008, 06:13 AM
ice_syncer
tangents
. If all the sides of a parallelogram touch a circle, show that the parallelogram is a rhombus.
• Jul 8th 2008, 06:34 AM
Plato
Quote:

Originally Posted by ice_syncer
. If all the sides of a parallelogram toucha circle, show that the parallelogram is a rhombus.

If by ‘touch’ you mean ‘tangent to’ then note that two tangents from an exterior point to a circle make congruent line segments to the circle. Think about opposite vertices.
• Jul 8th 2008, 06:44 AM
ice_syncer
Quote:

Originally Posted by Plato
If by ‘touch’ you mean ‘tangent to’ then note that two tangents from an exterior point to a circle make congruent line segments to the circle. Think about opposite vertices.

Yes I know thank you, I got it just after I posted it, we need to add the opposite sides and substitute it with the other equal tangents, I wish I knew how to delete posts..
• Jul 8th 2008, 07:49 AM
Soroban
Hello, ice_syncer!

I think this is what you meant . . .

Quote:

If all the sides of a parallelogram touch a circle,
show that the parallelogram is a rhombus.

Code:

              A      E                  B                 ∆ - * o * - - - - - - - ∆               /*          *          /             H o              *      /             *                *    /             /                      /           /*                  *  /           / *        * O      * /         /  *                  */         /                      /       /    *                *       /      *              o F     /          *          */     ∆ - - - - - - - * o * - ∆   D                  G      C
We have parallelogram $ABCD,$ inscribed circle $O$,
. . and points of tangency $E,F,G,H.$

Theorem: tangents from an external point to a circle are equal.

Hence, we have:
. . ${\color{blue}[1]}\;\;AE \:=\:AH$
. . ${\color{blue}[2]}\;\;BE \:=\:BF$
. . ${\color{blue}[3]}\;\;CG \:=\:CF$
. . ${\color{blue}[4]}\;\;DG \:=\:DH$

Add [1] and [4]: . $AE + DG \:=\:AH + DH \quad\Rightarrow\quad AE + DG \:=\:AD\;\;{\color{blue}[5]}$

Add [2] and [3]: . $BE + CG \:=\:BF + CF \quad\Rightarrow\quad BE + CG \:=\:BC\;\;{\color{blue}[6]}$

Add [5] and [6]: . $(AE+BE) + (DG + CG) \:=\:AD + BC$

So we have: . $AB + CD \:=\:AD + BC\;\;{\color{blue}[7]}$

Opposite sides of a parallelogram are equal: . $AB = CD,\;\;AD = BC$

So [7] becomes: . $2AB \:=\:2AD \quad\Rightarrow\quad AB \:=\:AD$

Since adjacent sides of the parallelogram are equal, all sides are equal.

Therefore, $ABCD$ is a rhombus.