Question

The triangle ABC has sides a = 13, b = 14 and c = 15 as shown the figure Line N bisects angle B and crosses side b at P The distance form A to P is

• A. $3\sqrt{7}$
• B. $7$
• C. $7\sqrt{2}$
• D. $7.5$

Answer 1

The correct option is D $7.5$

PN bisects $\angle ABC$
thus, by basic proportionality theorem,
$\frac{AB}{AC}=\frac{AP}{PC}$
$\frac{15}{13}=\frac{AP}{14-AP}$
$210-15AP=13AP$
$AP=\frac{210}{28}$
$AP=7.5$