Question

Let ${\Gamma }_1$ and ${\Gamma }_2$ be two circles touching each other externally at R. Let $l_1$ be a line which is tangent to ${\Gamma }_2$ at P and passing through the center $O_1$ of ${\Gamma }_1$. Similarly, let $l_2$ be a line which is tangent to ${\Gamma }_2$ at Q and passing through the center $O_2$ of ${\Gamma }_2$. Suppose $l_1$ and $l_2$ are not parallel and interesct at K. If $KP=KQ$, prove that the triangle $PQR$ is equilateral.

Answer 1

Suppose that P and Q lie on the opposite sides of line joining $O_1$ and $O_2$.
By symmetry we may assume that the configuration is as shown in the figure below.
Then we have $KP>K{O}_1>KQ$ since $K{O}_1$ is the hypotenuse of triangle $KQ{O}_1$. This is a contradiction to the given assumption, and therefore P and Q lie on the same side of the line joining $O_1$ and $O_2$.
Since $KP=KQ$ it follows that K lies on the radical axis of the given circles, which is the common tangent at R.
$\therefore KP=KQ=KR$ and hence K is the cirumcenter of $\Delta PQR$.
On the other hand, $\Delta KQ{O}_1$ and $\Delta KR{O}_1$ are both right-angled triangles with $KQ=KR$ and $Q{O}_1=R{O}_1$, and hence the two triangles are congruent.
$\therefore QK{O}_1=RK{O}_1$,
So $K{O}_1$, and hence PK is perpendicular to QR. Similarly, QK is perpendicular to PR, So it follows that K is the orthocenter of $\Delta PQR$.
Hence we have that $\Delta PQR$ is equilateral.