1. In the isosceles triangle, AB=AC, D is on AB, E is on the extension line of AC, BD=CE, and DE intersects BC at point F, which proves that DF=EF.

Prove:

The passing point d is DG∑AC.

∫DC∨ communication

∴∠DGB=∠ACB

AB = AC

∴∠ABC=∠ACB

∴∠DGB=∠ABC

∴DB=DG

BD = CE

∴DG=CE

∫DC∨ communication

∴∠DGC=∠ECG

∫∠DFG =∠ Conventional Armed Forces in Europe

∴△DGF is equal to△△ ECF.

∴DF=EF

2. It is known that △ABC, AH⊥BC are in H, the angle c = 35, AB+BH=HC, and the angle B is found.

Solution: extend CB, take BE = AB, and connect AE.

BE = AB

∴∠AEB=∠EAB

∴∠ABC=2∠AEB

∫a b+ BH = CH

∴EH=CH

∵AH⊥BC,AH=AH

∴△AEH is equal to △ △ACH.

∴∠AEB=∠C

∴∠ABC=2∠C

∠∠C = 35

∴∠ABC=70

3. In △ABC, ∠B=2∠C, and AD divided by ∠BAC. Verification: AC=AB+BD.

certificate

Take point E on AC, so that AE = AB.

Advertising split ∠BAC

∴∠BAD=∠EAD

AB = AE，AD=AD

∴△BAD is equal to△△△ EAD.

∴BD=DE,∠AED=∠B

∠∠AED =∠EDC+∠C，∠B=2∠C

∴∠EDC=∠C

∴DE=CE

∴AC=AB+BD

4. Given an obtuse triangle ADB(≈D is an obtuse angle), extend BD to F, take any point C on DF, connect AF, EF bisects AD vertically, and AD bisects ∠BAC. Prove that ∠ caf = ∠ b.

certificate

∵EF divides AD vertically.

∴FA=FD

∴∠FAD=∠FDA

Advertising split ∠BAC

∴∠CAD=∠BAD

∠∠FAD =∠CAF+∠CAD，∠FDA=∠BAD+∠B

∴∠CAF=∠B