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Question
In the given figure, OA·OB = OC·OD, Prove that ΔAOD ∼ ΔCOB.
Sum
Solution
Given:
OA × OB = OC × OD
To Prove: ΔAOD ∼ ΔCOB
Proof: OA × OB = OC × OD
⇒ `("OA")/("OC") = ("OD")/("OB")` ...(i)
Now, In ΔAOD and ΔBOC
`("OA")/("OC") = ("OD")/("OB")` ...(From (i))
∠AOD = ∠BOC ...(vertically opposite ∠s)
∴ ΔAOD ∼ ΔBOC ...(SAS similarity)
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