Question

In triangle $ABC,AL$ bisects angle $A$ and $CM$ bisects angle $C$. Points $L$ and $M$ are on $BC$ and $AB$, respectively. The sides of triangle $ABC$ are $a,b$ and $c$. Then $\frac{\overline{AM}}{\overline{MB}}=l\frac{\overline{CL}}{\overline{LB}}$, where $k$ is

• A. $1$
• B. $\frac{bc}{c^2}$
• C. $\frac{a^2}{bc}$
• D. $\frac{c}{b}$
• E. $\frac{c}{a}$

Answer 1

Answer: (E)

$\frac{c}{a}$

Draw $EG\perp AB.\frac{\overline{AD}}{\overline{AE}}=\frac{2}{3}=\frac{\overline{DF}}{\overline{EG}}$;
$\therefore \overline{EG}=\frac{3}{2}\overline{DF}=\frac{3\sqrt{2}}{4}$
Altitude $CDF=2\overline{EG}=\frac{3\sqrt{2}}{2}$,
$\therefore Area=\frac{1}{2}\cdot \sqrt{2}\cdot \frac{3\sqrt{2}}{2}=\frac{3}{2}$
or $x^2+x^2=2,x=1;y^2+x^2+{\left(\frac{x}{2}\right)}^2=1+\frac{1}{4}=\frac{5}{4}$.
Altitude $CDF=\sqrt{(2y{)}^2-{\left(\frac{\sqrt{2}}{2}\right)}^2}$
$=\sqrt{4\cdot \frac{5}{4}-\frac{1}{2}}=\sqrt{\frac{9}{2}}=\frac{3}{\sqrt{2}}=\frac{3\sqrt{2}}{2}$, etc.
The bisection of an angle of a triangle divides the opposite sides into segments proportional to the other two sides.
$△\frac{\overline{AM}}{\overline{MB}}=\frac{b}{a}$ and $\frac{\overline{CL}}{\overline{LB}}=\frac{b}{c}$. Since $\frac{c}{a}\cdot \frac{b}{c}=\frac{b}{a},k=\frac{c}{a}$.