AC+CB=AB
=>CB=5-2=3(cm)
Vì \(\frac{AC}{EB}<>\frac{AD}{CB}\)
nên ΔACD~ΔBEC
=>\(\hat{ACD}<>\hat{BEC}\)
mà \(\hat{BEC}=90^0-\hat{BCE}\)
nên \(\hat{ACD}<>90^0-\hat{BCE}\)
=>\(\hat{ACD}+\hat{BCE}<>90^0\)
Ta có: \(\hat{ACD}+\hat{BCE}+\hat{DCE}=180^0\)
=>\(\hat{DCE}=180^0-\left(\hat{DCA}+\hat{ECB}\right)\)
=>\(\hat{DCE}<>180^0-90^0=90^0\)
=>\(\hat{DCE}\) không là góc vuông


