Hard cyclic quadrilateral related question

156 Views Asked by At

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

I am expected to prove results using only Euclidean geometry, but I don't have any idea where to start. The result form $|PL|^2 = |PM| \cdot |PN|$ reminds me of the Secant-Tangent Theorem, but I can't solve it.

Any help would be appreciated

1

There are 1 best solutions below

0
On BEST ANSWER

Circle ABP has CA and CB as tangents.

Construct the circles ANPL and PMBL because the points are concyclic.

enter image description here

Then, $\angle 1 = \angle 2 =\angle 3 =\angle 4$.

Similarly, the red coded angles are equal. Therefore, $\triangle NLP \sim \triangle LMP$.

Result follows by setting up the ratios.