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.
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The sum of all triangles equals to 180 degrees
If we consider the first triangle which is ADC we will get
angle DAC + angle ACD + angle ADC = 180 Degrees
Angle DAC + 39 + 90 = 180 Degrees
Angle DAC = 180 - 39 - 90
Angle DAC = 51 Degrees
And if we consider triangle BEC we get
Angle BEC + angle ECB + angle BEC = 180
Angle EBC +39 +90= 180
Angle EBC = 180 - 39 -90
Angle BEC = 51 Degrees
If we consider the first triangle which is ADC we will get
angle DAC + angle ACD + angle ADC = 180 Degrees
Angle DAC + 39 + 90 = 180 Degrees
Angle DAC = 180 - 39 - 90
Angle DAC = 51 Degrees
And if we consider triangle BEC we get
Angle BEC + angle ECB + angle BEC = 180
Angle EBC +39 +90= 180
Angle EBC = 180 - 39 -90
Angle BEC = 51 Degrees
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