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Question
In the following figure, ∠ACB = ∠AMN. Prove that ΔAMN ∼ ΔАСВ.

Sum
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Solution
Given:
∠ACB = ∠AMN
Also, since B lies on AM and C lies on AN, the angle at A is common:
∠MAN = ∠CAB
Thus, in triangles △AMN and △ACB,
- ∠AMN = ∠ACB
- ∠MAN = ∠CAB
Therefore, by AA similarity,
△AMN ∼ △ACB
Hence proved.
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