What is the length of AR in the given triangle ABC?

Question

I need your help with a question on triangles. It is from the entrance exam of an institute in India for students in 10th grade.

In the given triangle ABC, if
DP, MQ AR and ES are perpendiculars to BC such that DP = 6 cm, MQ = 4 cm and ES = 8 cm
Then find AR.

The image of the figure

The image of the original question with options

MY ATTEMPT:
I tried to use Pythagoras theorem and tried substituting Pythagorean triplets to find the base.
Then I also tried similarity of triangles △DPC & △MQC and △ESB &△MQB to find the base BC.
I even tried to join AM and then extend it to intersect BC at F so as to use the Menelaus theorem and Ceva theorem. However, that didn't work for me too.

Any help would mean a lot to me.

Answer 1

I am going to solve using a combination of Euclidean Geometry and Analytic Geometry. Let $Q = (0,0)$ and let $B = (x_1, 0), P=(x_2,0), S = (x_3, 0), C = (x_4, 0)$ then by similar triangles

$ \dfrac{x_4}{ x_4 - x_2 } = \dfrac{2}{3} \tag{1}$

$ \dfrac{ -x_1}{x_3 - x_1} = \dfrac{1}{2} \tag{2}$

Now, let vertex $A = (x, y)$ , then again by similar triangles,

$ \dfrac{x_4 - x_3 }{x_4 - x} = \dfrac{8}{y} \tag{3}$

$ \dfrac{x_2 - x_1}{x - x_1} = \dfrac{6}{y} \tag{4}$

Taking the reciprocal of the last two equations, and re-arranging,

$ y = \dfrac{8 (x_4 - x)}{x_4 - x_3} = \dfrac{6(x - x_1)}{x_2 - x_1} \tag{5} $

Now looking back at the first two equations, we have

$ 3 x_4 = 2 x_4 - 2 x_2 \tag{6}$

So $x_4 = -2 x_2 \tag{7}$

And we also have

$ - 2 x_1 = x_3 - x_1\tag{8} $

So $ x_1 = - x_3 \tag{9}$

Substituting $(7)$ and $(9)$ into $(5)$, and writing $x_3 $ and $x_4$ in terms of $x_1 $ and $x_2$, gives us

$ y = \dfrac{ 8 (- 2 x_2 - x )}{- 2 x_2 + x_1} = \dfrac{ 6 (x - x_1)}{ x_2 - x_1} \tag{10}$

Solving for $x$ from the right most equation,

$$ 4 (-2 x_2 - x) (x_2 - x_1) = 3 (x - x_1) ( -2 x_2 + x_1) $$

Hence,

$$ x \left( 4 (x_1 - x_2) - 3 (-2 x_2 + x_1) \right) = 8 x_2 (x_2 - x_1) - 3 x_1 (-2 x_2 + x_1) $$

That is,

$$ x \left( x_1 + 2 x_2 \right) = 8 x_2^2 - 8 x_1 x_2 + 6 x_1 x_2 - 3 x_1^2 $$

$$ x \left( x_1 + 2 x_2 \right) = - 3 x_1^2 - 2 x_1 x_2 + 8 x_2^2 = (4x_2 - 3x_1) (2 x_2 + x_1)$$

Therefore,

$$ x = 4 x_2 - 3 x_1 $$

Substitute this in $(10)$

$$ y = 6\dfrac{ x - x_1}{x_2 - x_1} = 6 \left(\dfrac{ 4 (x_2 - x_1) }{x_2 - x_1} \right)= 24 $$

Therefore, $AR = 24$

Answer 2

Like you thought, this is a question about similar triangles, so you should write down whatever ratios you know. You've implied a few of them already, and those are the important ones. You could be really close to a solution, but are just missing one important observation.
Unforuntately, this isn't a question about nice Pythagorean triples like you thought. Pythagorean theorem might be helpful, but it won't be my first choice.

The hint I gave was

Let $AM$ intersect $BC$ at $F$. What is $\frac{ FM}{FA} $?

If we knew the answer to the hint, then using similar triangles, we can obtain $ \frac{FM}{FA} = \frac{ MQ}{AR} = \frac{4}{AR}$. So it remains to find $ \frac{ FM}{FA}$.

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Recall the well-known fact

$$\frac{FM}{FA} + \frac{ EM}{EB} + \frac{DM}{DC} = 1.$$

Hence, conclude that $ \frac{ FM}{FA} = \frac{1}{6} $ and so $AR = 24$.

Note

• The well-known fact can be proved using areas.
• Given that OP quoted Menelaus and Ceva's Theorem, I believe that they have seen this fact before.

Answer 3

Menelaus's theorem with line $\stackrel{\longleftrightarrow}{CMD}$ and triangle $\stackrel{\triangle}{ABE}$:

$$\frac{AC}{CE}\frac{EM}{MB} \frac{BD}{DA}=1$$ $$\frac{AR}{ES}\frac{ES-MQ}{MQ} \frac{DP}{AR-DP}=1$$ $$\frac x8\frac{8-4}4\frac6{x-6}=1$$ $$\frac{3x}{4x-24}=1$$ $$x=24$$