A triangle has corners points A, B, and C. Side AB has a length of $9$ $9$ . The distance between the intersection of point A's angle bisector with side BC and point B is $6$ $6$ . If side AC has a length of $14$ $14$ , what is the length of side BC?

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Answer 1 · 2017-06-10T19:51:09

The length of $B C = \frac{46}{3} = 15.3$ $BC=46/3=15.3$

Let $X$ $X$ be the point of intersection of the bisector with $B C$ $BC$

We apply the sine rule to triangle $A B X$ $ABX$

$\frac{A B}{sin ˆ A X B} = \frac{B X}{sin ˆ B A X}$ $(AB)/(sin hat(AXB))=(BX)/sin hat(BAX)$

$\frac{9}{sin ˆ A X B} = \frac{6}{sin ˆ B A X}$ $9/(sin hat(AXB))=6/sin hat(BAX)$ ....... $(1)$ $(1)$

Then, we apply the sine rule to triangle $A C X$ $ACX$

$\frac{A C}{sin ˆ A X C} = \frac{X C}{sin ˆ C A X}$ $(AC)/(sin hat(AXC))=(XC)/sin hat(CAX)$

$\frac{14}{sin ˆ A X C} = \frac{X C}{sin ˆ C A X}$ $(14)/(sin hat(AXC))=(XC)/sin hat(CAX)$ ......... $(2)$ $(2)$

Combining equations $(1)$ $(1)$ and $(2)$ $(2)$

$ˆ B A X = ˆ C A X$ $hat(BAX)=hat(CAX)$

And

$sin ˆ A X B = sin ˆ A X C$ $sin hat(AXB)=sin hat(AXC)$ as they are supplementary angles

$sin (π - θ) = sin θ$ $sin(pi-theta)=sin theta$

$\frac{9}{6} = \frac{14}{X C}$ $9/6=14/(XC)$

$X C = \frac{14 \cdot 6}{9} = \frac{28}{3}$ $XC=(14*6)/9=28/3$

Therefore,

$B C = B X + X C = 6 + \frac{28}{3} = \frac{46}{3} = 15.3$ $BC=BX+XC=6+28/3=46/3=15.3$

A triangle has corners points A, B, and C. Side AB has a length of 99 . The distance between the intersection of point A's angle bisector with side BC and point B is 66 . If side AC has a length of 1414 , what is the length of side BC?