Use properties of the bisector to find measures Geometry

According to the bisector theorem, we have:

\dfrac{AB}{AC}=\dfrac{BM}{CM}

Thus, we can write:

\dfrac{8}{7}=\dfrac{3}{CM}

CM=\dfrac{7 \times 3}{8}=\dfrac{21}{8}

According to the bisector theorem, we have:

\dfrac{AB}{AC}=\dfrac{BM}{CM}

Thus, we can write:

\dfrac{7}{AC}=\dfrac{4}{3}

AC=\dfrac{7 \times 3}{4}=\dfrac{21}{4}

According to the bisector theorem, we have:

\dfrac{AB}{AC}=\dfrac{BM}{CM}

Thus, we can write:

\dfrac{6}{AC}=\dfrac{5}{4}

AC=\dfrac{6 \times 4}{5}=\dfrac{24}{5}

According to the bisector theorem, we have:

\dfrac{AB}{AC}=\dfrac{BM}{CM}

Thus, we can write:

\dfrac{AB}{AC}=\dfrac{3}{3}=1

Hence:

AB = AC

We can conclude that ABC is an isosceles triangle. So:

\widehat{B} = \widehat{C}

Since \overline{AM} is a bisector, we have:

\widehat{b} = \widehat{a} = 20^\circ

Therefore:

\widehat{A} = 40^\circ

\widehat{A}+\widehat{B}+\widehat{C} = 180^\circ

40^\circ + \widehat{B} + \widehat{B} = 180^\circ

2\widehat{B} = 140^\circ

\widehat{B} = 70^\circ

According to the bisector theorem, we have:

\dfrac{AB}{AC}=\dfrac{BM}{CM}

Thus, we can write:

\dfrac{9}{5}=\dfrac{BM}{3}

BM=\dfrac{9 \times 3}{5}=\dfrac{27}{5}

According to the bisector theorem, we have:

\dfrac{AB}{AC}=\dfrac{BM}{CM}

Thus, we can write:

\dfrac{9}{4}=\dfrac{BM}{5}

BM=\dfrac{9 \times 5}{4}=\dfrac{45}{4}

According to the bisector theorem, we have:

\dfrac{AB}{AC}=\dfrac{BM}{CM}

Thus, we can write:

\dfrac{5}{AC}=\dfrac{4}{6}

AC=\dfrac{5 \times 6}{4}=\dfrac{30}4