\(a)CM:\Delta BHD\sim\Delta CKD\)
Xét \(\Delta BHD\) và \(\Delta CKD\) có:
\(\widehat{BHD}=\widehat{CKD}=90^0\)
\(\widehat{HDB}=\widehat{KDC}\) ( đối đỉnh)
Do đó: \(\Delta BHD\sim\Delta CKD\)
b) Xét \(\Delta ABH\) và \(\Delta ACK\) có:
\(\widehat{BAH}=\widehat{CAK}\left(gt\right)\)
\(\widehat{BHA}=\widehat{CKA}\left(=90^0\right)\)
Do đó: \(\Delta ABH\sim\Delta ACK\left(g-g\right)\)