Question

In the figure D, E, F are midpoints of sides AB, BC and AC respectively.P is the foot of the perpendicular from A to Side BC. Show that D, F, E, P are concylic ​

Answer 1

Answer:

Since ∠APC is a right angle, triangle APC is inscribed in a semicircle with diameter AC.  Therefore AF = FC = PF as these are all radii.

Hence triangle CPF is isosceles, so ∠CPF = ∠ FCP.

As D, E, F are midpoints, triangles DEF and CFE are congruent (SSS), so ∠FCE = ∠EDF.

Hence ∠EDF = ∠FCE = ∠FCP = ∠CPF = ∠EPF.

Since ∠EDF = ∠EPF, it follows that DFEP is cyclic.