Because AE

exist

It is an angular bisector, so it can be proved that AF=EF=BF.

So EF is the neutral line.

Namely: EF= 1/2(AD+BC)

And because AB=AF+BF=2EF.

So AB=AD+BC

Method 2: extend the switching between AE and BC to F.

Because in ∨ BC

Automatic exposure device

exist

It's an angular bisector, so AB=BF.

BE is the vertical line in the middle.

△ ade△ ECF can be proved according to the corner theorem.

So AD=CF

therefore

AB=BF=BC+CF=BC+AD

2、

Take a little e from BC, so that CE=AC connects DE.

According to the conditions, it can be proved that △ DEC △ DAC

So DE=AD

∠A=∠CED

Because BC=AC+AD.

So BE=DE

So ∠B = BDE∠CED =∠b+∠BDE.

So ∠ A = 2 ∠ B.

3, according to the known, available:

∠CAM+∠ACM=90

∠CDM+∠DCM=90

∠CAM=∠BAM

∠ADB=∠CDM

∠B+∠BAM+∠ADB=π

So ∠B=∠ACM+∠DCM