Question

In triangle ABC AP bisects angle BAC and AC>AB.prove that PC>BP

Answer 1

$\textbf{Concept:}$

$\textbf{Angle bisector theorem:}$

$\text{When vertical angle of a triangle is bisected, the bisector divides}$

$\text{the base into two segments which have the ratio as the order of other two sides.}$

$\textbf{Given: AC>AB }$

$\implies\;\frac{AB}{AC}<1......(1)$

$\text{In \triangle ABC, by angle bisector theorem we have}$

$\frac{BP}{PC}=\frac{AB}{AC}$

$\text{Using (1)}$

$\frac{BP}{PC}=\frac{AB}{AC}<1$

$\implies\frac{BP}{PC}<1$

$\implies\;BP<PC$

$\text{That is,}\bf\;PC>BP$

Find more:

AD=15 and DC=20. If BD is the bisector of angle ABC, what is the perimeter of the triangle ABC?

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