Ta có: ΔABC vuông tại A
=>\(\hat{ABC}+\hat{ACB}=90^0\)
=>\(\hat{ACB}=90^0-50^0=40^0\)
ΔECD vuông tại E
=>\(\hat{ECD}+\hat{EDC}=90^0\)
=>\(\hat{ECD}=90^0-40^0=50^0\)
Ta có; \(\hat{ACB}+\hat{BCD}+\hat{DCE}=180^0\)
=>\(\hat{BCD}=180^0-50^0-40^0=90^0\)
=>BC⊥CD
