Sixth question fast
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To prove: AB < AC
Proof: In ΔABC
∠B = 90o
Now,
∠A + ∠B + ∠C = 180o
∠A + ∠C = 90o
Therefore,
∠C must be acute angle or ∠C < ∠B
AB < AC (Sides opposite to the larger angle is larger)
Or, AB is shortest distance.
Proof: In ΔABC
∠B = 90o
Now,
∠A + ∠B + ∠C = 180o
∠A + ∠C = 90o
Therefore,
∠C must be acute angle or ∠C < ∠B
AB < AC (Sides opposite to the larger angle is larger)
Or, AB is shortest distance.
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