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Question
In the given figure, ∠1 = ∠2 and `(AC)/(BD)=(CB)/(CE)` Prove that Δ ACB ~ Δ DCE.
Solution
We have :
`(AC)/(BD)=(CB)/(CE)`
⟹ `(AC)/(CB)=(CD)/(CE)` (𝑆𝑖𝑛𝑐𝑒,𝐵𝐷=𝐷𝐶 𝑎𝑠 ∠1= ∠2 )
Also, ∠1= ∠2
i.e, ∠𝐷𝐵𝐶=∠𝐴𝐶𝐵
Therefore, by SAS similarity theorem, we get :
Δ ACB - Δ DCE
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