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In isosceles trapezoid ABCD, ad is parallel to BC, ab = CD, P is a point on BC, PE is parallel to CD, AC intersects with E, PF is parallel to AB, BD intersects with F
The parallel line of PC through f intersects DC with G
Because pf / / DC, G is on DC, so PF / / GC
So the quadrilateral fgcp is a parallelogram
So FG = PC (1), FP = GC (2)
Because this is an isosceles trapezoid, angle DBC = angle ACB
And FG / / PC and P on BC, so FG / / BC, so angle DFG = angle DBC = angle ACB (3)
Because EP / / AB, the angle EPC = angle ABC, and the two base angles in isosceles trapezoid are equal, that is, the angle ABC = angle DCB is equal to the angle EPC
Because of FG / / BC, angle DGF = angle DCB
Since angle DCB is also equal to angle EPC, angle DGF = angle EPC (4)
In triangle DGF and triangle EPC, there are (1), (3), (4), that is to say, the two angles of two triangles and the corresponding sides of them are equal, so the two triangles are congruent
Then EP = DG, FP = GC, GC + DG = DC in parallelogram fgcp
So PE + pf = DC, this is an isosceles trapezoid, ab = DC, so PE + pf = ab
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