a/ Xét \(ΔABC\) và \(ΔHBA\):
\(\widehat{BAC}=\widehat{BHA}(=90^\circ)\)
\(\widehat{B}:chung\)
\(→ΔABC\backsim ΔHBA(g-g)\)
\(→\dfrac{AB}{HB}=\dfrac{BC}{BA}\)
\(↔AB^2=HB.BC\)
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b/ Áp dụng định lý Pytago vào \(ΔABC\) vuông tại \(A\)
\(→BC=\sqrt{AB^2+AC^2}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10(cm)\)
\(ΔABC\backsim ΔHBA\)
\(→\dfrac{BC}{CA}=\dfrac{BA}{AH}\) hay \(\dfrac{10}{8}=\dfrac{6}{AH}\)
\(↔AH=4,8(cm)\)
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c/ \(DE\) là đường phân giác \(\widehat{ADB}\)
\(→\dfrac{EA}{EB}=\dfrac{DA}{DB}\)
\(→\dfrac{DB}{DC}.\dfrac{EA}{EB}.\dfrac{FC}{FA}=\dfrac{DB}{DC}.\dfrac{DA}{DB}.\dfrac{FC}{FA}=\dfrac{DA}{DC}.\dfrac{FC}{FA}\)
\(DF\) là đường phân giác \(\widehat{ADC}\)
\(→\dfrac{FC}{FA}=\dfrac{DC}{DA}\)
\(→\dfrac{DA}{DC}.\dfrac{DC}{DA}=1\)
\(→\) ĐPCM