Let
be a triangle, and let
be a point inside it such that
. The perpendiculars from
to
and
meet these lines at
and
, respectively, and
is the midpoint of
. Prove that
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Let $ABC$ be a triangle, and let $P$ be a point inside it such that $\angle PAC = \angle PBC$. The perpendiculars from $P$ to $BC$ and $CA$ meet these lines at $L$ and $M$, respectively, and $D$ is the midpoint of $AB$. Prove that $DL = DM.$
Školjka