In ABCD, ab = AC, CB = CD, point P is a point on diagonal AC, PE is perpendicular to BC and E, PF is perpendicular to CD and F, PE = PF is proved

In ABCD, ab = AC, CB = CD, point P is a point on diagonal AC, PE is perpendicular to BC and E, PF is perpendicular to CD and F, PE = PF is proved

Connecting BD, angle abd = angle ADB, angle CBD = angle CDB, therefore, angle B = angle D,
Triangle ABC is equal to ADC,
Angle ACD = angle ACB,
Triangle PFC is equal to PEC,
PE=PF