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!!

• 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 $90$° – enzotib May 14 '17 at 15:58

3 Answers

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,

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

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

• how did you get from the first ratio $\frac{BD}{AD}$=$\frac{AD}{CD}$=$\frac{AB}{CA}$ to $– Parley May 14 '17 at 16:49 • @Parley You mean $$\frac{BD}{CD}=\frac{BD}{AD}\cdot\frac{AD}{CD}=\left(\frac{AB}{CA}\right)^2$$? – CY Aries May 14 '17 at 16:51 • Yes! I mean from$\frac{BD}{AD}$=$\frac{AD}{CD}$=$\frac{AB}{CA}$to$\frac{BD}{CD}$=$\frac{BD}{AD}\frac{AD}{CD}$=$(\frac{AB}{CA})^{2}$– Parley May 14 '17 at 16:55 • Both$\frac{BD}{AD}$and$\frac{AD}{CD}$equal to$\frac{AB}{CA}$, as suggested by the similar triangle. – CY Aries May 14 '17 at 16:58 • Okay thanks! also, where does the$\frac{BD}{CD}$come from in that relationship? sorry this step really confused me! – 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):$BD/CD$=$(AB/AC)^2$. Altogether:$BM/CN$=$(AB/AC)^3\$.