As shown in the figure, AC ⊥ BC, ad ⊥ BD, ad = BC, CE ⊥ AB, DF ⊥ AB, perpendicular feet are e and f respectively, so, CE = DF?

As shown in the figure, AC ⊥ BC, ad ⊥ BD, ad = BC, CE ⊥ AB, DF ⊥ AB, perpendicular feet are e and f respectively, so, CE = DF?

Reason: in RT △ ABC and RT △ bad, ad = bcab = BA ≌ RT △ ABC ≌ RT △ bad (HL), ≌ AC = BD, ≌ cab = ≌ DBA. In △ ace and △ BDF, ≌ cab = ≌ DBA ≌ AEC = ≌ BFD = 90 ° AC = BD ≌ ace ≌ BDF (AAS), ≌ CE = DF

As shown in the figure, AC ⊥ BC, ad ⊥ BD, ad = BC, CE ⊥ AB, DF ⊥ AB, perpendicular feet are e and f respectively, so, CE = DF?

Reason: in RT △ ABC and RT △ bad, ad = bcab = BA ≌ RT △ ABC ≌ RT △ bad (HL), ≌ AC = BD, ≌ cab = ≌ DBA. In △ ace and △ BDF, ≌ cab = ≌ DBA ≌ AEC = ≌ BFD = 90 ° AC = BD ≌ ace ≌ BDF (AAS), ≌ CE = DF

It is known that: as shown in the figure, AC bisects ∠ bad, CE ⊥ AB in E & nbsp; CF ⊥ ad in F, and BC = DC

It is proved that ∵ AC bisects ∠ bad, CE ⊥ AB in E & nbsp; CF ⊥ ad in F, ∵ f = ∠ CEB = 90 ° CE = cf. in RT △ CEB and RT △ CFD, BC = DCCE = CF, ≌ CEB ≌ △ CFD (HL), ≌ be = DF

As shown in the figure, the AC bisector angle bad, CE, AB in E, CF, ad in F, and BC = DC are known. Try to judge the quantitative relationship between be and DF, and give reasons

Be = DF for the following reasons:
∵ AC bisecting angle bad, CE, AB in E, CF, ad in F
{CE = CF (a point on the bisector of the angle, the distance to both sides of the angle is equal)
∴∠CEB=∠CFD=90°
In △ CEB and △ CFD
{BC = DC (known)
CE = CF (proved)
∴△CEB≌△CFD（HL)
Be = DF (the corresponding sides of congruent triangles are equal)