Geometry Right Triangle and Circle 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!!
 A: 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,
$$\frac{BM\cdot AB}{CN\cdot CA}=\left(\frac{AB}{CA}\right)^4$$
$$\frac{BM}{CN}=\left(\frac{AB}{AC}\right)^3$$
A: 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.
A: $\angle AMD = 90 °$, 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: 
$BM/CN$ = $(AB/AC)^3$.
