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!!

  • 2
    $\begingroup$ 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

As $\triangle ABD\sim\triangle CAD$,




and thus


Using power of a point,

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


  • $\begingroup$ 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

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.






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.


$\angle AMD = 90 °$, 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):

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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.