Question 93
Let $ABCD$ be a convex quadrilateral. Let the diagonals $AC$ and $BD$ intersect in $P$. Let $PE$, $PF$, $PG$ and $PH$ be the altitudes from $P$ onto the sides $AB$, $BC$, $CD$ and $DA$ respectively. Show that $ABCD$ has an incircle if and only if
$$
\frac{1}{PE} + \frac{1}{PG} = \frac{1}{PF} + \frac{1}{PH}
$$
$$
\frac{1}{PE} + \frac{1}{PG} = \frac{1}{PF} + \frac{1}{PH}
$$