1. ## tangents

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

2. 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.

3. 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..

4. Hello, ice_syncer!

I think this is what you meant . . .

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.