Altitude Ratio

Since, in $\bigtriangleup ABC$ , the angular-bisector of $\angle A$ meets $BC$ at $N$ , from the Angle-Bisector Theorem, we have, $\frac{CN}{NB}=\frac{AC}{AB}=15$ .

Again, since, $\angle A=90^\circ$ and $AM\perp BC$ , we have, $\bigtriangleup AMC\sim\bigtriangleup AMB$ .

So, $\frac{CM}{AM}=\frac{AM}{MB}=\frac{AC}{AB}=15$

$\Rightarrow$ $CM=15.AM$ , and, $MB=\frac{AM}{15}$ .

Therefore, $\frac{CM}{MB}=\frac{15.AM}{\frac{AM}{15}}=15^2=225$

From Angle Bisector Theorem, we know that $\frac{AB}{NB}=\frac{AC}{CN}$ . This can be rearranged as $\frac{AC}{AB}=\frac{CN}{NB}=15$ . Because AM is an altitude drawn within a right triangle, we know that $\bigtriangleup ABC \sim \bigtriangleup MBA \sim \bigtriangleup MAC$ . We notice that $\bigtriangleup MBA$ and $\bigtriangleup MAC$ share $AM$ so we can relate $CM$ and $MB$ with the previous proportion since the triangles are similar. $\frac{CM}{MA}=\frac{AM}{MB}=15$ . From this, we get $\frac{CM}{MB}=225$ .

It's clear that angles $A$ and $B$ are acute angles. Let $CN = 15x$ and $BN = x$ , where $x$ is a positive real number.

Using Sine Rule on $\Delta ANB$ , we have $\frac{\sin \angle BAN}{BN} = \frac{\sin \angle B}{AN} \Rightarrow \frac{\sin \angle BAN}{x} = \frac{\sin \angle B}{AN} \Rightarrow \frac{\sin \angle BAN}{\sin \angle B} = \frac{x}{AN}$

Similarly, on $\Delta ANC$ , we have $\frac{\sin \angle CAN}{CN} = \frac{\sin \angle C}{AN} \Rightarrow \frac{\sin \angle CAN}{15x} = \frac{\sin \angle C}{AN} \Rightarrow \frac{\sin \angle CAN}{15 \sin \angle C} = \frac{x}{AN}$

Using the fact that the sine of an angle equals the cosine of its complement, i.e. $\sin (\frac{\pi}{2} - x) = \cos x$ , we get $\sin \angle C = \cos \angle B$ . Also, it's given that $\angle CAN = \angle BAN \Rightarrow \sin \angle CAN = \sin \angle BAN$ , from the angle bisector defintion. Hence, $\frac{\sin \angle CAN}{15 \cos \angle B} = \frac{x}{AN}$ . Equating the terms equal to $\frac{x}{AN}$ ,

$\sin \angle B = 15 \cos \angle B \Rightarrow \tan \angle B = 15$

Now, it's given that $AM$ is perpendicular to $BC$ . Then, it's clear that $\Delta AMB$ and $\Delta AMC$ are right triangles with right angle at $M$ .

Thus, $\tan \angle C = \frac{AM}{CM}$ and $\tan \angle B = \frac{AM}{BM}$ .

Thus, it suffices to find the value of $\displaystyle \frac{\tan \angle B}{\tan \angle C}$ , since it's equivalent to $\displaystyle \frac{CM}{MB}$ .

We use the property $\tan (\frac{\pi}{2} - x) = \cot x = \frac{1}{\tan x}$ . Thus, $\frac{\tan \angle B}{\tan \angle C} = \tan^2 \angle B$ .

$\tan \angle B = 15 \Rightarrow \tan^2 \angle B = \boxed{225}$

From the Angle Bisector Theorem, CA/AB=15. Without loss of generality, assume that AB=1 and AC=15 to make computations simpler. Then BC=sqrt226. Let x=BM and y=AM. Then CM=sqrt226-x. From the Pythagorean Theorem, x^2+y^2=1 and y^2+(sqrt226-x)^2=225. Solving for x and y, we find the answer to be 225.

using sine rule, CN/sin(<CAN)=AN/sin(C)-----(1) BN/sin(<NAB)=AN/sin(B)-----(2) but <CAN=<NAB=45 <B+<C=90 so sinC=sin(90-B)=cosB--------(3) CN/BN=15--------(4) from (1),(2),(3) and (4), tanB=15

now let <CAM=x <MAB=90-x therefore <B=x (since <MAB+<AMB+<B=180) using sine rule, CM/sinx=AM/sinC BM/sin(90-x)=AM/sinB on solving, we get CM/BM=(sinB.sinx)/(sinC.cosx)=tan^2(B)=(15)^2=225

Let CN=15y and BN=y. Let AM=h. By the AngleBisector Theorem, we can let AC=15x and AB=x. By the Phytagorean Theorem: (15x)^2+x^2=(16y)^2 => 256/226=(x/y)^2. Because (AC) (AB)/2=(BC) h/2 (area of triangle), h=15/16 (x^2 y). By Phytagorean Theorem: (CM)/(MB)=sqrt((AC)^2-(AM)^2)/sqrt((AB)^2-(AM)^2)=sqrt(225x^2-(15/16 x^2/y)^2)/sqrt(x^2-(15/16 x^2/y)^2). Factoring it an x^2 => sqrt(225-225/256 (x/y)^2)/sqrt(1-225/256 (x/y)^2). Plugging in for (x/y)^2 gets you that the ratio is 225.

As the tags suggest, we'll need the angle bisector theorem , concept of similar figures, triangle properties and ... geometry (duh!).

Whenever you see a problem with angle bisectors of a triangle and ratios, your first approach should be the angle bisector theorem. This approach works (almost) all the time. So let's see what happens when we use the angle bisector theorem here.

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Since $AN$ is the the bisector of $\angle BAC$ , $\frac{CA}{BA}=\frac{CN}{NB}=21$ $\cdots (1)$ .

A little bit of angle chasing reveals to us that all the right triangles $\triangle ABC$ , $\triangle ABM$ and $\triangle ACM$ , are similar to each other. That means:

$\frac{CA}{BA}=\frac{CM}{AM}$ $\cdots (2)$ [comparing $\triangle ABC$ and $\triangle ACM$ ]

$\frac{CA}{BA}=\frac{AM}{MB}$ $\cdots (3)$ [comparing $\triangle ABC$ and $\triangle ABM$ ]

Multiply $(2)$ and $(3)$ to get ${(\frac{CA}{BA})}^2=\frac{CM}{MB}$ . From $(1)$ , $\frac{CA}{BA}=21$ .

So, $\frac{CM}{MB}=21^2=\boxed{441}$ and we're done!

By the angle bisector theorem, $\dfrac{CN}{NB}=\dfrac{CA}{AB}=21$ Then, by similar triangles, $\dfrac{CM}{MB}=\dfrac{CM}{MA}\cdot\dfrac{AM}{MB}=\dfrac{CA}{AB}\cdot\dfrac{CA}{AB}=21\cdot 21=\boxed{441}$