Đáp án:
$AB = AC = \dfrac{{25}}{2};BC = 15$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
BC = x\\
AC = AB = y
\end{array} \right.\left( {x,y > 0} \right)\\
+ ){S_{ABC}} = \dfrac{1}{2}AH.BC = \dfrac{1}{2}BK.AC\\
\Rightarrow 10.x = 12.y \Rightarrow x = \dfrac{6}{5}y\left( 1 \right)\\
+ )\Delta AHC;\widehat {AHC} = {90^0}\\
\Rightarrow A{H^2} + H{C^2} = A{C^2}\\
\Rightarrow {10^2} + {\left( {\dfrac{x}{2}} \right)^2} = {y^2}\left( 2 \right)\\
\left( 1 \right),\left( 2 \right) \Rightarrow \left\{ \begin{array}{l}
x = \dfrac{6}{5}y\\
{10^2} + {\left( {\dfrac{x}{2}} \right)^2} = {y^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{6}{5}y\\
100 + {\left( {\dfrac{3}{5}y} \right)^2} = {y^2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{6}{5}y\\
y = \dfrac{{25}}{2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 15\\
y = \dfrac{{25}}{2}
\end{array} \right.\\
\Rightarrow AB = AC = \dfrac{{25}}{2};BC = 15
\end{array}$
Vậy $AB = AC = \dfrac{{25}}{2};BC = 15$
