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 inscribed circle ,
. . and points of tangency
Theorem: tangents from an external point to a circle are equal.
Hence, we have:
. .
. .
. .
. .
Add [1] and [4]: .
Add [2] and [3]: .
Add [5] and [6]: .
So we have: .
Opposite sides of a parallelogram are equal: .
So [7] becomes: .
Since adjacent sides of the parallelogram are equal, all sides are equal.
Therefore, is a rhombus.