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प्रश्न
In the following figure, ABC is an equilateral triangle and P is any point in AC;
prove that: BP > PA
योग
उत्तर
In ΔABC,
AB = BC = CA ...[ ABC is an equilateral triangle ]
∴ ∠A = ∠B = ∠C
∴ ∠A = ∠B = ∠C = `(180°)/3`
In ΔABP,
∠A = 60°
∠ABP< 60°
∴ ∠A > ∠ABP
⇒ BP > PA ....[ Side opposite to greater side is greater ]
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Inequalities in a Triangle - If two sides of a triangle are unequal, the greater side has the greater angle opposite to it.
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