Sửa lại đề nhé: \(\dfrac{AH}{DH}=k\)
Do \(CF\perp AB;AD\perp BC\Rightarrow\) góc AFH = góc ADB
\(\Rightarrow\Delta AFH\sim\Delta ADB\left(g.g\right)\Rightarrow\)góc ABC = góc AHF = góc DHC
\(\Rightarrow tgB=tgD\widehat{H}C=\dfrac{DC}{DH}\)
lại có: tgC = \(\dfrac{AD}{DC}\)
\(\Rightarrow tgB.tgC=\dfrac{DC}{DH}.\dfrac{AD}{DC}=\dfrac{AD}{DH}=\dfrac{DH+AH}{DH}=1+k\)