08/13/2025
$\triangle{ABC}$ is inscribed in a circle and the tangent lines at $A,B$ meet at $M$, the tangent lines at $B,C$ meet at $N$. $BP\perp AC$ at $P$, show that $\angle{MPB}=\angle{NPB}$.
Prove 1:
\(\begin{multline}
\shoveleft \text{Let }E, F \text{ on extended }AC \text{ such that }EM\perp MN, FN \perp MN\\
\shoveleft \implies EMBP, FNBP \text{ is both cyclic }\\
\shoveleft \implies \angle{MPB}=\angle{MEB}, \angle{NBP}=\angle{NFB}\\
\shoveleft \text{Let }MB=MA=a, NB=NC=b\\
\shoveleft \angle{MAB}=\angle{MBA}=\angle{BCA}=\alpha, \angle{NBC}=\angle{NCB}=\angle{BAC}=\beta\\
\shoveleft \implies sin\angle{AEM}=sin\angle{NBP}=sin\angle{NFC}\\
\shoveleft \implies \angle{MAE}=180^{\circ}-\alpha-\beta=\angle{NCF} \implies sin\angle{MAE}=sin\angle{NCF}\\
\shoveleft \implies \dfrac{ME}{MB}=\dfrac{ME}{MA}=\dfrac{sin\angle{MAE}}{sin\angle{AEM}}=\dfrac{sin\angle{NCF}}{sin\angle{NFC}}=\dfrac{NF}{NC}=\dfrac{NF}{NB}\\
\shoveleft \implies \triangle{EMB}\sim\triangle{FNB}\implies \angle{MEB}=\angle{NFB}\implies \angle{MPB}=\angle{NPB} \blacksquare
\end{multline}\)
Prove 2:
\(\begin{multline}
\shoveleft \text{Let }E, F \text{ on extended }AC \text{ such that }EM\perp MN, FN \perp MN\\
\shoveleft \implies EMBP, FNBP \text{ is both cyclic }\\
\shoveleft \implies \angle{MPB}=\angle{MEB}, \angle{NBP}=\angle{NFB}\\
\shoveleft \text{Let }D \text{ on }EF \text{ such that }ND=NF\implies \angle{CDN}=180^{\circ}-\angle{NFD}\\
\shoveleft \implies \angle{MEA}=\angle{NBP}=180^{\circ}-\angle{NFD}=\angle{CDN}\\
\shoveleft \angle{MAB}=\angle{MBA}=\angle{BCA}, \angle{NBC}=\angle{NCB}=\angle{BAC}\\
\shoveleft \implies \angle{MAE}=\angle{NCF} \implies \triangle{AEM}\sim\triangle{CDN}\\
\shoveleft \implies \dfrac{ME}{MA}=\dfrac{ND}{NC}\implies \dfrac{ME}{MB}=\dfrac{NF}{NB} \implies \triangle{EMB}\sim\triangle{FNB}\\
\shoveleft \implies \angle{MEB}=\angle{NFB} \implies \angle{MPB}=\angle{NPB} \blacksquare
\end{multline}\)
Prove 3:
\(\begin{multline}
\shoveleft \text{Let }O \text{ be the center of the circumcircle of }\triangle{ABC}\\
\shoveleft \implies OB\perp MN, MO\perp AB, NO \perp BC\\
\shoveleft \implies \angle{ABM}=\angle{ACB}=\angle{MOB}, \angle{CAB}=\angle{CBN}=\angle{NOB}\\
\shoveleft \implies \triangle{BMO}\sim\triangle{PBC}, \triangle{BNO}\sim\triangle{PBA}\\
\shoveleft \implies \dfrac{MB}{BO}=\dfrac{BP}{PC},\dfrac{NB}{BO}=\dfrac{BP}{PA}\implies \dfrac{MB}{NB}=\dfrac{PA}{PC}=\dfrac{MA}{NC}\\
\shoveleft \angle{PAM}=\angle{PCN}\implies \triangle{MAP}\sim\triangle{NCP}\\
\shoveleft \implies \angle{MPA}=\angle{NPC} \implies \angle{MPB}=\angle{NPB}\blacksquare
\end{multline}\)