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 $\displaystyle ABCD,$ inscribed circle $\displaystyle O$,
. . and points of tangency $\displaystyle E,F,G,H.$
Theorem: tangents from an external point to a circle are equal.
Hence, we have:
. . $\displaystyle {\color{blue}[1]}\;\;AE \:=\:AH$
. . $\displaystyle {\color{blue}[2]}\;\;BE \:=\:BF$
. . $\displaystyle {\color{blue}[3]}\;\;CG \:=\:CF$
. . $\displaystyle {\color{blue}[4]}\;\;DG \:=\:DH$
Add [1] and [4]: .$\displaystyle AE + DG \:=\:AH + DH \quad\Rightarrow\quad AE + DG \:=\:AD\;\;{\color{blue}[5]}$
Add [2] and [3]: .$\displaystyle BE + CG \:=\:BF + CF \quad\Rightarrow\quad BE + CG \:=\:BC\;\;{\color{blue}[6]}$
Add [5] and [6]: .$\displaystyle (AE+BE) + (DG + CG) \:=\:AD + BC$
So we have: .$\displaystyle AB + CD \:=\:AD + BC\;\;{\color{blue}[7]}$
Opposite sides of a parallelogram are equal: .$\displaystyle AB = CD,\;\;AD = BC$
So [7] becomes: .$\displaystyle 2AB \:=\:2AD \quad\Rightarrow\quad AB \:=\:AD$
Since adjacent sides of the parallelogram are equal, all sides are equal.
Therefore, $\displaystyle ABCD$ is a rhombus.