Geometry Problem MCQ

Question

$AD$ bisects the angle $A$ of $\triangle ABC$ and meets $BC$ at $D$ such that $BD/DC = AB/AC$. If $BC = k$, $CA = l$ and $AB = m$, then the length of $DC$ is: a) $$kl/(m+l)$$ b) $$k(m+l)/l$$ c) $$km/(m+l)$$ d) $$k(m+l)/m$$

I used the given equation to solve for $DC$ and took $DC$ as $x$ and $BC$ as $(k-x)$ and got the following quadratic equation:

$$x^2 + lx/m - lk/m = 0$$

whose root is

$$\frac{-l + \sqrt{l^2 + 4lmk}}{2m}$$

I am not able to solve from here. Can anyone suggest a way to proceed? Thanks.

Answer 1

That seems quite complicated. The answer that I got is option a, that is $\frac{kl}{m+l}$

I did this using the following diagram: Labelled Triangle Diagram

Now, the question give us that $\frac{BD}{DC}$=$\frac{AB}{AC}$

Using the variables mentioned in the question as well as the diagram above, we get: $$\frac xy=\frac ml$$

Multiplying both sides by $y$ as well as adding $y$ on both sides give us: $$x+y=\frac{my}{l} + y$$

Now, we bring back the assumption that $x+y=k$ and take $y$ common on the other side, which give us: $$k=y(\frac{m+l}{l})$$

Now, rearranging the formula gives us the following: $$DC=y=\frac{kl}{m+l}$$

Answer 2

The right answer is a), at it follows from the equation

$$\frac{k-x}x = \frac{m}l$$

which is nothing else as the rewriting of the given relation

$$\frac {BD} {DC} = \frac {AB} {AC}$$

as  $BD = BC - DC \space(= k - x$).

Answer 3

$$\frac{BD}{AB}=\frac{DC}{AC}=\frac{BD+DC}{AB+AC} \Rightarrow \frac{x}{l}=\frac{k}{m+l} \Rightarrow x=\frac{lk}{m+l}$$

The first term equality is a standard property of proportion.