As shown in the figure, ad bisects ∠ BAC, de ⊥ AB in E, DF ⊥ AC in F, and DB = DC, proving: EB = FC

As shown in the figure, ad bisects ∠ BAC, de ⊥ AB in E, DF ⊥ AC in F, and DB = DC, proving: EB = FC

It is proved that: ∵ ad bisects ∠ BAC, de ⊥ AB in E, DF ⊥ AC in F, ∵ de = DF; ∵ de ⊥ AB in E, DF ⊥ AC in F. ∵ in RT △ DBE and RT △ DCF, de = dfdb = DC ≌ RT △ DBE ≌ RT △ DCF (HL); ≌ EB = FC