Bài làm:
Ta có: \(\Delta CDE~\Delta CAB\left(g.g\right)\)
vì: \(\hept{\begin{cases}\widehat{CDE}=\widehat{CAB}=90^0\\\widehat{ECD}=\widehat{BAC}\left(chung\right)\end{cases}}\)
\(\Rightarrow\frac{CD}{CA}=\frac{CE}{CB}\left(1\right)\)
Xét 2 tam giác: \(\Delta BEC\)và \(\Delta ADC\)có:
\(\hept{\begin{cases}\frac{CD}{CA}=\frac{CE}{CB}\left(1\right)\\\widehat{BCE}=\widehat{ACD}\left(chung\right)\end{cases}}\)
\(\Rightarrow\Delta BEC~\DeltaÂDC\left(c.g.c\right)\)
=> đpcm
Học tốt!!!!