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Question

In $$\triangle ABC,\angle ACB=90^{o} $$ Seg $$CD \bot$$ side $$AB$$ and seg $$DE \bot$$ side $$CB$$ show that $$CD^{2}\times AC=AD\times AB\times DE$$
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Solution

We have,

ABC is a right angle triangle.

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

And $$CD\bot AB\,and\,DE\bot AB$$

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:-

$$In\,\Delta ACB\,and\,\Delta ADC$$

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

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

Thus, By AA similarity.

Then,

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

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

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

Now, Similarly

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

Now,

$$In\,\Delta CED\,and\,\Delta CDB$$

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

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

 

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,

$$\Delta ADC\sim \Delta CED$$

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

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

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.


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