Question

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

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.Maths

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