a) Do \(AD\) là phân giác \(\widehat{A}\). Theo tính chất tia phân giác ta có:
\(\dfrac{DB}{AB}=\dfrac{DC}{AC}=\dfrac{DB+DC}{AB+AC}=\dfrac{30}{20+25}=\dfrac{30}{55}=\dfrac{6}{11}\)
\(\Rightarrow\left\{{}\begin{matrix}BD=\dfrac{6}{11}.AB=\dfrac{120}{11}\left(cm\right)\\DC=\dfrac{6}{11}.AC=\dfrac{150}{11}\left(cm\right)\end{matrix}\right.\)
b) Xét \(\Delta BHD\) và \(\Delta CKD\) có:
\(\widehat{BHD}=\widehat{CKD}=90^0\) (gt)
\(\widehat{BDH}=\widehat{CDK}\) (hai góc đối đỉnh)
\(\Rightarrow\Delta BHD\sim\Delta CKD\) (g.g)
c) Do \(\Delta BHD\sim\Delta CKD\) nên \(\dfrac{HD}{KD}=\dfrac{BH}{CK}\)
Xét \(\Delta ABH\) và \(\Delta ACK\) có:
\(\widehat{AHB}=\widehat{AKC}=90^0\) (gt)
\(\widehat{BAH}=\widehat{CAK}\) (Do AD là tia phân giác)
\(\Rightarrow\Delta ABH\sim\Delta ACK\) (g.g) \(\Rightarrow\dfrac{AH}{AK}=\dfrac{BH}{CK}\)
Vậy \(\dfrac{AH}{AK}=\dfrac{HD}{KD}\Rightarrow AH.KD=AK.HD\)
