Let $N$ be the midpoint of arc $ABC$ of the circumcircle of $\Delta ABC$, and $NP$, $NT$ be the tangents to the incircle of this triangle. The lines $BP$ and $BT$ meet the circumcircle for the second time at points $P_1$ and $T_1$ respectively. Prove that $PP_1 = TT_1$.
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Tags: geometry
AlastorMoody
06.03.2019 16:18
Just show that $BPINT$ is a cyclic pentagon, hence, $BP_1, BT_1$ are isogonal $\implies$ $AC||P_1T_1$ finish off with congruency after that
Pluto1708
06.03.2019 16:50
Easy to observe that $N$ is the center of spiral similarity taking $PT \to P_1T_1$ but since $NP=NT$ thus the triangles are congruent and the result follows
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P10.pdf (117kb)
anantmudgal09
06.03.2019 19:51
Another favourite Note that $\angle IBN=\angle IPN=\angle ITN=90^{\circ}$ so $BPITN$ is cyclic. Now $IP=IT$ proves that $BP$ and $BT$ are isogonal. Thus, $NP_1=NT_1$. Now $NP=NT$ and $\angle PNT=\angle PBT=\angle P_1BT_1=\angle P_1NT_1$ so $\angle PNP_1=\angle TNT_1$. Thus $\triangle NPP_1 \cong \triangle NTT_1$ and $PP_1=TT_1$.