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ABCD is a rectangle. Δ ADE and Δ ABF are two triangles such that ∠E=∠F as shown in the figure. Prove that AD x AF=AE x AB.
Ans: Consider Δ ADE and Δ ABF
∠D = ∠B = 90o
∠E = ∠F (given)
∴Δ ADE ≅ Δ ABF
AD/AB = AE/AF
⇒ AD x AF = AB x AE Proved
1+2cos(2x=0
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01010011 01100101 01101101 01110000 01100101 01110010 00100000 01000110 01101001 00100001
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