As shown in the figure, in RT △ ABC, the bisector of ∠ a = 90 °, the bisector of ∠ B intersects AC at e, the height ah on the side of intersecting BC is at D, DF / / BC intersects AC at F through D, and the verification is: AE = FC

As shown in the figure, in RT △ ABC, the bisector of ∠ a = 90 °, the bisector of ∠ B intersects AC at e, the height ah on the side of intersecting BC is at D, DF / / BC intersects AC at F through D, and the verification is: AE = FC

prove:
Eg ⊥ HC at point E and eg ∥ ah at point G
AE = eg is obtained from ∠ Abe = ∠ GBE ∠ BAE = ∠ Bge = 90 & # 186
∵ ∠ADE=∠BDH=90º-∠DBH ∠AEB=90º-∠ABE
∵ ∠ABE=∠DBH
∴ ∠ADE=∠AEB ∴AD=AE ∴ AD=AE =EG
And ∵ DF ∥ DC ∥ AFD = ∥ C ∥ ADF = ∥ EGC = 90 ∥ 186; ad = eg
∴⊿ADF≌⊿EGC ∴AF=EC ∴AE=FC