Problem Section #3 a) Circles $O_1$ and $O_2$ interest at two points $B$ and $C$, and $BC$ is the diameter of circle $O_1$. Construct a tangent line of circle $O_1$ at $C$ and intersecting circle $O_2$ at another point $A$. Join $AB$ to intersect circle $O_1$ at point $E$, then join $CE$ and extend it to intersect circle $O_2$ at point $F$. Assume $H$ is an arbitrary point on line segment $AF$. Join $HE$ and extend it to intersect circle $O_1$ at point $G$, and then join $BG$ and extend it to intersect the extend of $AC$ at point $D$. Prove: $\frac{AH}{HF}=\frac{AC}{CD}$.
Problem
Source: Nepal Mathematics Olympiad
Tags: geometry
Sphr
22.04.2018 15:17
Easy question By easy angle chasing we have $HFGC$ is cyclic. We call it $W_1$. Let the intersection of this circle with segment $AC$ (Not point $C$) be $H'$. By the PoP of the point $A$ for circle $W_1$ we obtain $HH'$||$FC$. So we have: $$\frac{AH}{HF}=\frac{AH'}{CH'}$$So now we should prove that: $$\frac{AH'}{CH'}=\frac{AC}{CD}$$thus $$\frac{AH'}{AC}=\frac{AC}{AD}$$Thus $AC^2=AH'.AD$ But we know that $AC^2=AE.AB$. So we have to prove $$AB.AE=AD.AH'$$And this requires to prove $\triangle ABD$ ~ $\triangle AH'E$. By ($HH'$||$FC$) we have $\angle H'HE=\angle CEG=\angle CBG$ Thus $\angle CDG=\angle AEH'$
Math-Ninja
22.04.2018 15:21
Sphr wrote: Easy question By easy angle chasing we have $HFGC$ is cyclic. We call it $w_1$ Let the intersection of this circle with segment $AC$ (Not point C) be $H'$. By the PoP of the point $A$ for circle $w_1$ we obtain $HH'$ is parallel to $FC$. So we have: $$\frac{AH}{HF}=\frac{AH'}{CH'}$$So now we should prove that: $$\frac{AH'}{CH'}=\frac{AC}{CD}$$thus $$\frac{AH'}{AC}=\frac{AC}{AD}$$Thus $AC^2=AH'.AD$ But we know that $AC^2=AE.AB$. So we have to prove $$AB.AE=AD.AH'$$And this requires to prove $\triangle ABD~\triangle AH'E$ By ($HH'$ is parallel to $FC$) we have $\angle H'HE=\angle CEG=\angle CBG$ Thus $\angle CDG=\angle AEH'$ FTFY
Satyamkumar
28.08.2020 08:12
AH'/H'C=AH/HF IS NOT THE POWER OF POIN THIS IS SIMPLE ANGLE CHASING RESULT WHICH STATES HH'is parallel to FC
SomeUser221104
28.08.2020 09:06
Simple trig bash Let $\angle AEH = \alpha$, we have $\angle FAE = \angle FAB = \angle FCB = \angle BAC$, thus $$\frac{AH}{HF} = \frac{EA \sin \angle AEH}{FE \sin \angle HEF }=\frac{EA}{FE} \tan \alpha = \tan \alpha \cot \angle FAE = \tan \alpha \cot \angle BAC$$On the other hand, we have $\angle DBC = \angle GEC = \angle HEF = \frac{\pi}{2} - \alpha $. $$\frac{AC}{CD} = \frac{AC}{BC \tan \angle DBC}= \frac{AC}{BC} \tan \alpha = \cot \angle BAC \tan \alpha $$Thus we have : $$\frac{AH}{HF} = \cot \angle BAC \tan \alpha = \frac{AC}{CD} $$