Ta có:`AD=AE+DE`
`⇒17=8+DE`
`⇒DE=17-8`
`⇒DE=9(cm)`
Ta có:`(AB)/(DE)=6/9=2/3`
`(AE)/(CD)=8/12=2/3`
`⇒(AB)/(DE)=(AE)/(CD)=2/3`
Xét `ΔABE` và `ΔDEC` có:
`(AB)/(DE)=(AE)/(CD)=2/3(cmt)`
`hat{BAE}=hat{EDC}=90^o`
`⇒ΔABE=ΔDEC(c.g.c)`
`⇒hat{E_1}=hat{C_1}(2` góc tương ứng `)`
Mà `hat{E_2}+hat{C_1}=90^o(2` góc phụ nhau `)`
`⇒hat{E_2}+hat{E_1}=90^o`
Ta có:`hat{E_1}+hat{BEC}+hat{E_2}=180^o`
`⇒(hat{E_1}+hat{E_2})+hat{BEC}=180^o`
`⇒90^o +hat{BEC}=180^o`
`⇒hat{BEC}=180^o-90^o`
`⇒hat{BEC}=90^o(đpcm)`