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प्रश्न
In the following figure, ABC is an equilateral triangle and P is any point in AC;
prove that: BP > PC
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उत्तर
In ΔBPC,
∠C = 60°
∠CBP < 60°
∴∠C > ∠CBP
⇒ BP > PC ....[ Side opposite to greater side is greater ]
संबंधित प्रश्न
In triangle ABC, AB > AC and D is a point inside BC.
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(iii) AE > AD
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prove that: ∠ADC is greater than ∠ADB.
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Prove that: C > A.
In quadrilateral ABCD, side AB is the longest and side DC is the shortest.
Prove that: D > B.
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Write the angles xo, yo and zo in ascending order of their values.
In the following diagram; AD = AB and AE bisect angle A. 
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In a quadrilateral ABCD; prove that:
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