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Question
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.
Sum
Solution
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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