AB = AC

∴∠ABC=∠ACB

∫∠BAC = 45

∴∠abc=∠acb=( 180-45)/2=67.5

∫△a 1b 1C?△ABC，

∴∠b 1a 1c=∠bac=45,∠a 1b 1c=∠abc=67.5,bc=b 1c

∴∠BB 1C=∠ABC=67.5

∴∠ab 1a 1= 180-∠bb 1c-∠a 1b 1c= 180-67.5-67.5=45

∴∠ab 1a 1=∠b 1a 1c

∴AB//A 1C

2、

△A 1AB and△ CBA1congruence.

Prove:

AB = AC，AC=A 1C

∴AB=A 1C

∵AB//A 1C

∴ parallelogram ABCA 1

∴AA 1=BC

∫a 1B = ba 1

∴△A 1AB and△ CBA1congruence.

3. Solution:

∫△A2B2C?△ABC、

∴∠A2B2C=∠ABC=67.5

∴∠ab2a2= 180-∠a2b2c= 180-67.5= 1 12.5