In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $\overline{AC}$. If $BD = 2CD$,prove that $AC = 5CD$.
(N/A) $1$. In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{BD} \perp \overline{AC}$.
$2$. By the property of geometric mean in a right-angled triangle,we have $BD^2 = AD \cdot CD$.
$3$. Given $BD = 2CD$,substitute this into the equation: $(2CD)^2 = AD \cdot CD$.
$4$. This simplifies to $4CD^2 = AD \cdot CD$.
$5$. Dividing both sides by $CD$ (since $CD \neq 0$),we get $AD = 4CD$.
$6$. Since $AC = AD + CD$,substitute $AD = 4CD$ into the equation: $AC = 4CD + CD$.
$7$. Therefore,$AC = 5CD$.
$2$. By the property of geometric mean in a right-angled triangle,we have $BD^2 = AD \cdot CD$.
$3$. Given $BD = 2CD$,substitute this into the equation: $(2CD)^2 = AD \cdot CD$.
$4$. This simplifies to $4CD^2 = AD \cdot CD$.
$5$. Dividing both sides by $CD$ (since $CD \neq 0$),we get $AD = 4CD$.
$6$. Since $AC = AD + CD$,substitute $AD = 4CD$ into the equation: $AC = 4CD + CD$.
$7$. Therefore,$AC = 5CD$.
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