Giải thích các bước giải:

 a) Ta có:

$\begin{array}{l}
\Delta AEH \sim \Delta BDH\left( {g.g} \right)\\
 \Rightarrow \dfrac{{HA}}{{HB}} = \dfrac{{HE}}{{HD}}
\end{array}$

Khi đó

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {AHB} = \widehat {EHD}\left( {dd} \right)\\
\dfrac{{HA}}{{HB}} = \dfrac{{HE}}{{HD}}
\end{array} \right.\\
 \Rightarrow \Delta AHB \sim \Delta EHD\left( {c.g.c} \right)\\
 \Rightarrow \dfrac{{HA}}{{HE}} = \dfrac{{AB}}{{ED}}\\
 \Rightarrow HA.ED = HE.AB
\end{array}$

c) Ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {BDH} = \widehat {ADC} = {90^0}\\
\widehat {DBH} = \widehat {DAC}\left( { + \widehat {ACD} = {{90}^0}} \right)
\end{array} \right.\\
 \Rightarrow \Delta BDH \sim \Delta ADC\left( {g.g} \right)\\
 \Rightarrow \dfrac{{DB}}{{DA}} = \dfrac{{DH}}{{DC}}\\
 \Rightarrow \dfrac{{DB}}{{DH}} = \dfrac{{DA}}{{DC}}
\end{array}$

Khi đó:

$\begin{array}{l}
\Delta ADC;\widehat D = {90^0}\\
 \Rightarrow DC = \sqrt {A{C^2} - A{D^2}}  = 4cm\\
 \Rightarrow \dfrac{{DB}}{{DH}} = \dfrac{{DA}}{{DC}} = \dfrac{3}{4}
\end{array}$

Đáp án:

Giải thích các bước giải:

a) Ta có:

ΔAEH∼ΔBDH(g.g)⇒HAHB=HEHDΔAEH∼ΔBDH(g.g)⇒HAHB=HEHD

Khi đó

⎧⎪⎨⎪⎩ˆAHB=ˆEHD(dd)HAHB=HEHD⇒ΔAHB∼ΔEHD(c.g.c)⇒HAHE=ABED⇒HA.ED=HE.AB{AHB^=EHD^(dd)HAHB=HEHD⇒ΔAHB∼ΔEHD(c.g.c)⇒HAHE=ABED⇒HA.ED=HE.AB

c) Ta có:

⎧⎨⎩ˆBDH=ˆADC=900ˆDBH=ˆDAC(+ˆACD=900)⇒ΔBDH∼ΔADC(g.g)⇒DBDA=DHDC⇒DBDH=DADC{BDH^=ADC^=900DBH^=DAC^(+ACD^=900)⇒ΔBDH∼ΔADC(g.g)⇒DBDA=DHDC⇒DBDH=DADC

Khi đó:

ΔADC;ˆD=900⇒DC=√AC2−AD2=4cm⇒DBDH=DADC=34