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Let AD be the altitude corresponding to the hypotenuse BC of the right triangle ABC. The circle of diameter AD intersects AB at M and AC at N shown. Prove $\frac{BM}{CN}$= $\bigg(\frac{AB}{AC}\bigg)^{3}$.

So far I have...

The power of B is $BD^{2}=(BM)(BA)$

The power of C is $CD^{2}=(CN)(CA)$

I am stuck after this. Any help would be appreciated!!

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    $\begingroup$ the graphics representation is a little bit wrong, because $M$, $N$ and $C_1$ should be aligned. This is due to the fact that $\hat{BAC}$ on the graphics is greater than $90$° $\endgroup$ – enzotib May 14 '17 at 15:58
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As $\triangle ABD\sim\triangle CAD$,

$$\frac{BD}{AD}=\frac{AD}{CD}=\frac{AB}{CA}$$

Therefore,

$$\frac{BD}{CD}=\frac{BD}{AD}\cdot\frac{AD}{CD}=\left(\frac{AB}{CA}\right)^2$$

and thus

$$\frac{BD^2}{CD^2}=\left(\frac{AB}{CA}\right)^4$$

Using power of a point,

$$\frac{BM\cdot AB}{CN\cdot CA}=\left(\frac{AB}{CA}\right)^4$$

$$\frac{BM}{CN}=\left(\frac{AB}{AC}\right)^3$$

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  • $\begingroup$ how did you get from the first ratio $\frac{BD}{AD}$=$\frac{AD}{CD}$=$\frac{AB}{CA}$ to $ $\endgroup$ – Parley May 14 '17 at 16:49
  • $\begingroup$ @Parley You mean $$\frac{BD}{CD}=\frac{BD}{AD}\cdot\frac{AD}{CD}=\left(\frac{AB}{CA}\right)^2$$? $\endgroup$ – CY Aries May 14 '17 at 16:51
  • $\begingroup$ Yes! I mean from $\frac{BD}{AD}$=$\frac{AD}{CD}$=$\frac{AB}{CA}$ to $\frac{BD}{CD}$=$\frac{BD}{AD}$ $\frac{AD}{CD}$=$(\frac{AB}{CA})^{2}$ $\endgroup$ – Parley May 14 '17 at 16:55
  • $\begingroup$ Both $\frac{BD}{AD}$ and $\frac{AD}{CD}$ equal to $\frac{AB}{CA}$, as suggested by the similar triangle. $\endgroup$ – CY Aries May 14 '17 at 16:58
  • $\begingroup$ Okay thanks! also, where does the $\frac{BD}{CD}$ come from in that relationship? sorry this step really confused me! $\endgroup$ – Parley May 14 '17 at 17:03
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Hint: What can you say about triangle $BMD$ given that A forms a right angle? Try to express $\frac{BM}{CN}$ in terms of the relations of sides that you know.

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Solution: The angle in M of $AMD$ is right, since it sees a diameter of the circle. Hence $BMD$ has two angles equal to $ABC$, and is similar to it. The same holds for $AMD$ and $DNC$. Also, the quadrilateral $AMDN$ has four right angles, and therefore it is a rectangle, and $AM=DN$. Now we operate. $$\frac{BM}{CN} = \frac{BM}{MD}\frac{MD}{AM}\frac{AM}{DN}\frac{DN}{CN} = \frac{AB}{BC}\frac{AB}{BC}1\frac{AB}{BC}$$ so we get the desired result.

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$\angle AMD = 90 °$, so $MD$ || $AC$.

(Thales circle over $AD$.)

Intercept Theorem:

1) $BM/BA$ = $BD/BC$ .

$\angle AND = 90°$, so $ND$ || $AB$.

(Thales circle over $AD$.)

Intercept Theorem:

2) $CN/CA$ = $CD/CB$.

Dividing1) by 2):

$BM/CN$ × $CA/BA$ = $BD/CD$ ;

$BM/CN$ = $BA/CA$ × $BD/CD$.

To express the ratio $BD/CD$ we look at similar triangles.

$\triangle ADC$ is similar to $\triangle CAD$ is similar to $\triangle CAB$:

Let $\angle DAB$ = $\angle BCA$ = $\beta$.

1) $AB/AC$ = $tan(\beta)$.

2) $AD/DC$ = $tan(\beta)$.

3) $BD/AD$ = $tan(\beta)$.

Multiplying: 2) × 3) :

$BD/CD$ = $[tan(\beta)]^2$.

Hence using 1):

$BD/CD$ = $(AB/AC)^2$.

Altogether:

$BM/CN$ = $(AB/AC)^3$.

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