A triangle has corners at points A, B, and C. Side AB has a length of $15$ $15$ . The distance between the intersection of point A's angle bisector with side BC and point B is $7$ $7$ . If side AC has a length of $21$ $21$ , what is the length of side BC?

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Answer 1 · 2016-03-17T09:46:41

$B C = 16.8$ $BC=16.8$

Let the figure be as follows, in which $A P$ $AP$ bisects angle $A$ $A$ and cuts $B C$ $BC$ at $P$ $P$ .

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As $A P$ $AP$ bisects $B C$ $BC$ , according to angle bisector theorem

$\frac{A B}{A C} = \frac{B P}{B C}$ $(AB)/(AC)=(BP)/(BC)$ i.e. here we have

$\frac{15}{21} = \frac{7}{C P}$ $15/21=7/(CP)$ and $C P = 21 \times \frac{7}{15} = 9.8$ $CP=21xx7/15=9.8$

Hence, $B C = B P + C P = 7 + 9.8 = 16.8$ $BC=BP+CP=7+9.8=16.8$ .

A triangle has corners at points A, B, and C. Side AB has a length of 1515 . The distance between the intersection of point A's angle bisector with side BC and point B is 77 . If side AC has a length of 2121 , what is the length of side BC?