Draw AF⊥BC and EG⊥BCar∆ABCar∆EBC = 12×BC×AF12×BC×EG = AGEG 1In ∆AFD and ∆EGD∠AFD = ∠EGD 90° each∠ADF = ∠EDG Common⇒∆AFD~∆EGD AA⇒AFEG = ADED = FDGD Corresponding sides of similar ∆'s are proportional⇒AFEG = ADEDnow, from 1, we getar∆ABCar∆EBC = ADED

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Draw AF ⊥ BC and EG ⊥ BC \frac{ar (∆ ABC)}{ar (∆ EBC)} = \frac{\frac{1}{2} \times BC \times AF}{\frac{1}{2} \times BC \times EG} = \frac{AG}{EG} . . . . . (1) In ∆ AFD and ∆ EGD ∠ AFD = ∠ EGD (90 ° each) ∠ ADF = ∠ EDG (Common) \Rightarrow ∆ AFD ~ ∆ EGD (AA) \Rightarrow \frac{AF}{EG} = \frac{AD}{ED} = \frac{FD}{GD} (Corresponding sides of similar ∆' s are proportional) \Rightarrow \frac{AF}{EG} = \frac{AD}{ED} now, from (1), we get \frac{ar (∆ ABC)}{ar (∆ EBC)} = \frac{AD}{ED}

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