Show that if a circle can be inscribed in quadrilateral $1$ ($AESH$) and in quadrilateral $2$ ($KCLS$), then circle can be inscribed also in the quadrilateral $ABCD$.
Here is a picture:
The places where inscribed circle is tangent to quadrilateral $1$ are marked as red points (blue points for places where inscribed circle is tangent to quadrilateral $2$).
As you see I've already took into account same lengths of some sides somming form the fact of inscribed circles (lengths: $a, b, c, d, e, f, g, h$). Then I think that I need to use the fact that both circles are also inscribed in bigger tiangles:
- circle that is inscribed in quadrilateral $1$ is also inscribed in triangles $AEG$ and $AFH$
- circle that is inscribed in quadrilateral $2$ is also inscribed in triangles $SKG$ and $SFL$
From that I got $4$ equations:
- $n+g = m + h$
- $m+h + e + c = o + p + d$
- $l+g = k + f$
- $k + f + e + c = b + i + j$
But I don't know what to do next. Ultimately I need to prove that: $AB + DC = AD + BC$.
