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प्रश्न
The sides AB and AC of a triangle ABC are produced; and the bisectors of the external angles at B and C meet at P. Prove that if AB > AC, then PC > PB.
बेरीज
उत्तर
In ΔABC,
AB > AC,
⇒ ∠ACB < ∠ABC
⇒ 180° - ∠ACB < 180° - ∠ABC
∠BCE < ∠DBC
`("∠BCE")/2 < ("∠DBC") /2`
∠BCP < ∠CBP
In ΔBCP
∠BCP < ∠CBP
⇒ BP < CP
⇒ PC > PB
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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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संबंधित प्रश्न
Prove that the straight line joining the vertex of an isosceles triangle to any point in the base is smaller than either of the equal sides of the triangle.
In the following figure, ABC is an equilateral triangle and P is any point in AC;
prove that: BP > PA
From the following figure; 
prove that:
(i) AB > BD
(ii) AC > CD
(iii) AB + AC > BC.
