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Known: as shown in the figure, in the square ABCD, AE ⊥ BF, the perpendicular foot is p, AE and CD intersect at point E, BF and ad intersect at point F, verification: AE = BF
It is proved that the ∵ quadrilateral ABCD is a square, AE ⊥ BF, ∵ DAE + ≌ AED = 90 °, ≌ DAE + ≌ AFB = 90 °, ∵ AED = ≌ AFB, and ∵ ad = AB, ≌ bad = ≌ D, ≌ AED ≌ ABF, ≌ AE = BF
In square ABCD, AE is perpendicular to BF, the perpendicular foot is p, AE and CD intersect at point E, BF and ad intersect at point F
If ad = 4, de = 3, find the length of PE
13 out of 5
Easy to prove triangle AED and triangle ABF congruence
So AF = de = 3, because the side length of the square is 4, so AE = BF = 5 can be obtained from Pythagorean theorem. Because AP is perpendicular to BF, so AP = 12 / 5 (inverted by area), so PE = 12 / 5 = 13 / 5
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