In Delta ABC,the bisector of angle A intersec | vedclass

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(A) According to the Angle Bisector Theorem,the bisector of an angle of a triangle divides the opposite side into segments that are proportional to th

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$BD = 4, DC = 5$

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(A) According to the Angle Bisector Theorem,the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.Therefore,$\frac{BD}{DC} = \frac{AB}{AC}$.Given $AB = 8$ and $AC = 10$,we have $\frac{BD}{DC} = \frac{8}{10} = \frac{4}{5}$.Let $BD = 4x$ and $DC = 5x$.Since $BC = BD + DC = 9$,we have $4x + 5x = 9$,which implies $9x = 9$,so $x = 1$.Thus,$BD = 4(1) = 4$ and $DC = 5(1) = 5$.

In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. If $AB = 8$,$AC = 10$ and $BC = 9$,find $BD$ and $DC$.

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