1     Let (\Gamma_1) and (\Gamma_2) be two circles touching each other externally at R. Let (O_1) and (O_2) be the centres of (\Gamma_1) and (\Gamma_2), respectively. Let (\ell_1) be a line which is tangent to (\Gamma_2) at P and passing through (O_1), and let (\ell_2) be the line tangent to (\Gamma_1) at Q and passing through (O_2). Let (K=\ell_1\cap \ell_2). If KP=KQ then prove that the triangle PQR is equilateral.


We note that (O_1 R O_2 ) is a straight line (why?)
Also (\Delta O_1 Q O_2 , \Delta O_1 P O_2 ) are right angled triangles with right angles at point Q and P respectively.
Hence (\Delta O_1 Q K , \Delta O_2 P K ) are similar (vertically opposite angles and right angles)
Thus ( \frac{KP}{KQ} = \frac{O_2 P} {O_1 Q} = 1) as KP =KQ.
Hence the radii of the two circles are equal..
This implies R is the midpoint of (O_1 O_2) hence the midpoint of hypotenuse of  (\Delta O_1 Q O_2)
(O_1 R = RQ = R O_2) since all are circum-radii of (\Delta O_1 Q O_2).
Hence  (\Delta O_1 Q R) is equilateral, similarly (\Delta O_2 P R) is also equilateral.
Thus (\angle PRQ) is (60^o) also \( RQ = O_1 R = O_2 R = RP \).
Hence triangle PQR is equilateral.