Ta có :
BACˆ=1800−ABCˆ−ACBˆ=600BAC^=1800-ABC^-ACB^=600
→DACˆ=BADˆ=12BACˆ=300DAC^=BAD^=12BAC^=300

→ADCˆ=BADˆ+ABDˆ=1100ADC^=BAD^+ABD^=1100

b.Ta có AB=AE, DABˆ=DAEˆ→ΔBAD=ΔAECAB=AE, DAB^=DAE^→∆BAD=∆AEC(c.g.c)

c.Từ câu b →BDEˆ=2ADBˆ=2(DACˆ+ACDˆ)=1400BDE^=2ADB^=2DAC^+ACD^=1400

Mà IBCˆ=12ABCˆ=400IBC^=12ABC^=400

→IBDˆ+BDEˆ=1800→DE//BI