Hello andy69 [IMG]file:///C:/Users/janet/AppData/Local/Temp/moz-screenshot-8.png[/IMG] Originally Posted by
andy69 In the attached diagram, $\displaystyle TK$ is a tangent at $\displaystyle P$ to the circle $\displaystyle C_1$. We have to prove that $\displaystyle \triangle PKL$ is isosceles.
Proof
$\displaystyle \angle LPK = \angle TPM$ (vertically opposite) $\displaystyle =\angle PQM$ (alternate segment)
$\displaystyle =\angle KLP$ (exterior angle of a cyclic quadrilateral)
$\displaystyle \therefore \triangle PKL$ is isosceles.
Grandad