Question

Let ABC be an acute-angled triangle with AC is not equal to BC and let O be the circumcenter and F the foot of altitude through C. Further, let X and Y be the feet of perpendiculars dropped from A and B to (the extension of) CO. The line FO intersects the circumcircle of triangle FXY, second time at P. Prove that OP is less than OF.

Answer 1

ow, we know that the sum of all angles of a triangle equals [math]180^{\circ} [/math].

First, by considering [math] \bigtriangleup ADC [/math], we get

[math] \angle DAC + \angle ACD + \angle ADC = 180^{\circ} [/math]

=> [math] \angle DAC + 39^{\circ} + 90^{\circ} = 180^{\circ} [/math]

=> [math] \angle DAC = 180^{\circ} - 39^{\circ} - 90^{\circ} [/math]

=> [math] \angle DAC = 51^{\circ} [/math]

Next, by considering [math]\bigtriangleup BEC[/math], we get

[math] \angle EBC+ \angle ECB + \angle BEC = 180^{\circ} [/math]

=> [math] \angle EBC + 39^{\circ} + 90^{\circ} = 180^{\circ} [/math]

=> [math] \angle EBC = 180^{\circ} - 39^{\circ} - 90^{\circ} [/math]

=> [math] \angle EBC = 51^{\circ} [/math]