Question

In $△ABC,\angle ACB={90}^o$ Seg $CD\perp$ side $AB$ and seg $DE\perp$ side $CB$ show that $C{D}^2\times AC=AD\times AB\times DE$

Answer 1

We have,

ABC is a right angle triangle.

In which $\angle ACB={90}^o$

And $CD\perp ABandDE\perp AB$

Where D is any point in AB and E is any point in CB.

Prove that:- $C{D}^2\times AC=AD\times AB\times DE$

Proof:-

$In\Delta ACBand\Delta ADC$

$\angle CAB\cong \angle DAC\left(\text{Reflexive}\right)$

$\angle ACB\cong \angle ADC\left(\text{Right}\text{angle}\right)$

Thus, By AA similarity.

Then,

$\Delta ACB\sim \Delta ADC......\left(1\right)$

$\Rightarrow \frac{AC}{AD}=\frac{AB}{AC}$

$\Rightarrow A{C}^2=AB\times AD......\left(2\right)$

Now, Similarly

$\Delta ACB\sim \Delta CDB......\left(3\right)$

Now,

$In\Delta CEDand\Delta CDB$

$\angle ECD\cong \angle DCB\left(\text{Reflexive}\right)$

$\angle CED\cong \angle CDB\left(\text{Right}\text{angle}\right)$

By AA similarity.

$\Delta CED\sim \Delta CDB.......\left(4\right)$

From equation (3) and (4) to,

$\Delta ACB\sim \Delta CED......\left(5\right)$

By equation (1) and (5) to, we get,

$\Delta ADC\sim \Delta CED$

$\frac{AC}{CD}=\frac{DC}{ED}$

$C{D}^2=AC\times DE......\left(6\right)$

From equation (2) to,

$A{C}^2=AB\times AD$

On multiplying both side by $C{D}^2$ and we get,

$C{D}^2\times A{C}^2=C{D}^2\times AB\times AD$

$C{D}^2\times A{C}^2=AC\times DE\times AB\times AD$

$C{D}^2\times AC=DE\times AB\times AD$

$C{D}^2\times AC=AD\times AB\times DE$

Hence proved.