can someone help in solving this triangle-circumcircle question?

Question

$AB$ is a chord of a circle and the tangents at $A$, $B$ meet at $C$. If $P$ is any point on the circle and $PL$, $PM$, $PN$ are the perpendiculars from $P$ to $AB$, $BC$, $CA$. Prove that $PL^2= PM\cdot PN$.

I tried solving for angles and proving $\triangle PLM$ and $\triangle PNL$ similar but was unable to do so.

Answer 1

Let $\Gamma$ denote the circle. The line $MB$ is tangent to $\Gamma$ at $B$, so $\angle PAB=\angle PBM$. Since $PLBM$ is a cyclic quadrilateral, $\angle PBM=\angle PLM$, whence $$\angle PAB=\angle PBM=\angle PLM\,.$$ Likewise, $PLAN$ is a cyclic quadrilateral, so $$\angle PNL=\angle PAL=\angle PAB\,.$$ Consequently, $$\angle PNL=\angle PLM\,.$$

Similarly, $NA$ is tangent to $\Gamma$ at $C$ so $\angle PBA=\angle PAN$. Since $PLBM$ is a cyclic quadrilateral, $\angle PML=\angle PBL=\angle PBA$, whence $$\angle PML=\angle PBA=\angle PAN\,.$$ Likewise, $PLAN$ is a cyclic quadrilateral, so $$\angle PLN=\angle PAN\,.$$ Consequently, $$\angle PLN=\angle PML\,.$$

Therefore, in the triangles $MPL$ and $LPN$, we have $\angle PLM=\angle PNL$ and $\angle PML=\angle PLN$. Thus, $MPL$ and $LPN$ are similar triangles, whence $$\frac{PM}{PL}=\frac{PL}{PN}\,,$$ as desired.

Answer 2

Consider two cases.

1. $P$ is located on the circle such that $P$ and $C$ are placed in two different sides respect to $AB$.

Since, $PNAL$ and $PLBM$ are cyclic, we obtain: $$\measuredangle NLP=\measuredangle NAP=\measuredangle ABP=\measuredangle LMP.$$ Also, $$\measuredangle NPL=180^{\circ}-\measuredangle NAL=180^{\circ}-\measuredangle MBL=\measuredangle MPL,$$ which gives $\Delta NLP\sim\Delta LMP.$

Id est, $$\frac{PL}{PM}=\frac{PN}{PL}$$ or $$PL^2=PM\cdot PN.$$ 2. $P$ is located on the circle such that $P$ and $C$ are placed in the same side respect to $AB$.

Since, $PNAL$ and $PLBM$ are cyclic, we obtain: $$\measuredangle NLP=\measuredangle NAP=\measuredangle ABP=\measuredangle LMP.$$ Also, $$\measuredangle NPL=180^{\circ}-\measuredangle NAL=180^{\circ}-\measuredangle MBL=\measuredangle MPL,$$ which gives $\Delta NLP\sim\Delta LMP.$

Id est, $$\frac{PL}{PM}=\frac{PN}{PL}$$ or $$PL^2=PM\cdot PN$$ again.

If $P\equiv A$ then $PL=PN=0$.

If $P\equiv B$ then $PL=PM=0$ and we obtain $PL^2=PM\cdot PN$ again.