Question
In the given figure, $\angle\text{ACB}=90^\circ$ and $\text{CD}\perp\text{AB}.$
Prove that $\frac{\text{BC}^2}{\text{AC}^2}=\frac{\text{BD}}{\text{AD}}.$
Prove that $\frac{\text{BC}^2}{\text{AC}^2}=\frac{\text{BD}}{\text{AD}}.$
✓
Answer
Given: $\angle\text{ACB}=90^\circ$ and $\text{CD}\perp\text{AB}$
To Prove: $\frac{\text{BC}^2}{\text{AC}^2}=\frac{\text{BD}}{\text{AD}}$
Proof:
In $\triangle\text{ACB}$ and $\triangle\text{CDB}$
$\angle\text{ACB}=\angle\text{CDB}=90^\circ$ (Given)
$\angle\text{ABC}=\angle\text{CBD}$ (common)
By AA similarity-criterion $\triangle\text{ACB}\sim\triangle\text{CDB}$
When two triangles are similar, the ratios of the lengths of their corresponding sides are proportional.
$\therefore\frac{\text{BC}}{\text{BD}}=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\text{BC}^2=\text{BD}.\text{AB}\ .....(1)$
In $\triangle\text{ACB}$ and $\triangle\text{ADC}$
$\angle\text{ACB}=\angle\text{ADC}=90^\circ$ (Given)
$\angle\text{CAB}=\angle\text{DAC}$ (common)
By AA similarity-criterion $\triangle\text{ACB}\sim\triangle\text{ADC}$
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
$\therefore\frac{\text{AC}}{\text{AD}}=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\text{AC}^2=\text{AD}.\text{AB}\ .....(2)$
Divinding (2) by (1), we get
$\frac{\text{BC}^2}{\text{AC}^2}=\frac{\text{BD}}{\text{AD}}$
To Prove: $\frac{\text{BC}^2}{\text{AC}^2}=\frac{\text{BD}}{\text{AD}}$
Proof:
In $\triangle\text{ACB}$ and $\triangle\text{CDB}$
$\angle\text{ACB}=\angle\text{CDB}=90^\circ$ (Given)
$\angle\text{ABC}=\angle\text{CBD}$ (common)
By AA similarity-criterion $\triangle\text{ACB}\sim\triangle\text{CDB}$
When two triangles are similar, the ratios of the lengths of their corresponding sides are proportional.
$\therefore\frac{\text{BC}}{\text{BD}}=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\text{BC}^2=\text{BD}.\text{AB}\ .....(1)$
In $\triangle\text{ACB}$ and $\triangle\text{ADC}$
$\angle\text{ACB}=\angle\text{ADC}=90^\circ$ (Given)
$\angle\text{CAB}=\angle\text{DAC}$ (common)
By AA similarity-criterion $\triangle\text{ACB}\sim\triangle\text{ADC}$
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
$\therefore\frac{\text{AC}}{\text{AD}}=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\text{AC}^2=\text{AD}.\text{AB}\ .....(2)$
Divinding (2) by (1), we get
$\frac{\text{BC}^2}{\text{AC}^2}=\frac{\text{BD}}{\text{AD}}$
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