Triangle $ABC$ has incircle $\omega$ and incenter $I$. On its sides $AB$ and $BC$ we pick points $P$ and $Q$ respectively, so that $PI = QI$ and $PB > QB$. Line segment $QI$ intersects $\omega$ in $T$. Draw a tangent line to $\omega$ passing through $T$; it intersects the sides $AB$ and $BC$ in $U$ and $V$ respectively. Prove that $PU = UV + VQ$!
Problem
Source:
Tags: geometry, incenter
jayme
27.01.2017 14:53
Dear Mathlinkers, I think that there is a little typo... PU + QV = UV Sincerely Jean-Louis
MemocanTs61
31.12.2020 22:10
Let $\omega$ touch the sides $AB$ and $AC$ at $D$ and $E$, respectively. We can see that $\angle IPD = \angle IQE$. $PI = QI$ and $ID=IE$. So triangles $IPD$ and $IQE$ are congruent. That means $PD=QE$. So, $$PU=PD+DU=QE+UT=QV+VE+UT=QV+VT+UT=QV+UV$$
trying_to_solve_br
03.08.2021 17:17
Cute problem mehhhhh Note that $UV+VQ=PU \Leftrightarrow VT + UT + VQ = PU \Leftrightarrow VD + UE + VQ = PU \Leftrightarrow VD + VQ = PE \Leftrightarrow QD = PE \Leftrightarrow QD^2=PE^2 \Leftrightarrow Pot_\gamma Q=Pot_\gamma P \Leftrightarrow IP=IQ$, which is true.