Let $ABC $ be a triangle and $D $ be the feet of $A $-altitude. $E,F $ are defined on segments $AD,BC $,respectively such that $\frac {AE}{DE}=\frac{BF}{CF} $. Assume that $G $ lies on $AF $ such that $BG\perp AF $.Prove that $EF $ is tangent to the circumcircle of $CFG $. Proposed by Mehdi Etesami Fard
Problem
Source: Iran RMM TST 2019, day1 p1
Tags: geometry, ratio
rocketscience
30.07.2019 21:25
Let the line through $E$ parallel to $BC$ meet $AF$ at a point $H$. Note that $\triangle AEH \sim \triangle BGF$, and also that $\frac{BF}{CF} = \frac{AE}{ED} = \frac{AH}{FH}$, so in fact $\triangle AEF \sim \triangle BGC$. Now from this follows $\angle FCG = \angle GFE$ so $EF$ is tangent to the circumcircle of $\triangle CFG$.
sbealing
14.03.2020 19:45
Let $X$ be a point on line $BG$ such that $\angle XCB=90^{\circ}$. Then $XCFG$ is cyclic. As $\angle AGB=\angle ADB=90^{\circ}$ we get $AGDB$ cyclic so $\angle DAF=\angle DBG=\angle CBX$ and hence $\triangle BXC \cap F \simeq \triangle AFD \cap E$ so: $$\angle EFG=\angle EFA=\angle FXG=\angle FCG$$Hence $EF$ is tangent to $\odot CFG$
Miku3D
27.04.2020 17:54
I got a third solution, cool
All angles used here are directed $\mod 180^{\circ}$
Let $S$ be the second intersection of $(ABD)$ and the circle with diameter $CD$
Clearly, $AD \perp DC$ implies that $AD$ is tangent to $(CSD)$
Claim: $\triangle SAD \sim \triangle SBC$
ProofWe have
$$\measuredangle SAD = \measuredangle SBD = \measuredangle SBC$$$$\measuredangle SDA = \measuredangle SCD = \measuredangle SCB$$So the two triangles are similar
Thus, the Claim is proven $\blacksquare$
Claim: Quadrilateral $CSGF$ is cyclic
ProofWe have
$$\measuredangle SCF = \measuredangle SCD = \measuredangle SDA = \measuredangle SGA = \measuredangle SGF$$Which implies that CSGF is cyclic
Thus, the Claim is proven $\blacksquare$
Claim: $\triangle SEF \sim \triangle SDC$
ProofSince $\frac{AE}{ED}=\frac{BF}{FC}$, then $\triangle SED \sim \triangle SFC$ as well
So $S$ is the center of the spiral similarity sending $ED$ to $FC$, so it is also the center of the spiral similarity sending $EF$ to $DC$, so $\triangle SEF \sim \triangle SDC$
Thus, the Claim is proven $\blacksquare$
Therefore, $$\measuredangle GFE = \measuredangle GFS + \measuredangle SFE$$$$=\measuredangle GCS + \measuredangle SCD$$$$= \measuredangle GCD$$$$=\measuredangle GCF$$Which implies that $EF$ is tangent to $(CFG)$ $\blacksquare$
Steve12345
02.02.2021 17:00
I got fourth and fifth Let $\frac {AE}{DE}=\frac{BF}{CF}=k$
Add a point on $BG$ such that $\frac{BP}{PG}=k$. Notice: $\triangle BGF \sim \triangle ADF$ so obviously $\triangle AEF \sim \triangle BPF$ hence $\angle PFB =\angle GFE$ but also notice $PF || GC$ $\rightarrow$ $\angle PFB =\angle GCB=\angle GFE$ and we're done
Add a point on ray $AF$ such that $\frac{AF}{FP}=k$.
Notice $EF || DP$ so $\angle GFE = \angle GPD$ so it's enough to prove $GDPC$ cyclic.
Luckily, this is easy since $\frac{BF}{CF}=\frac{AF}{FP}$ so $CP||AB$.
Now $\angle CPG = \angle CPA = \angle BAG = \angle GDF$ (by cyclic $BDGA$) so
$\angle GDC= \angle GPC$ and we're done.
Attachments:
IsoOnly
02.02.2021 20:33
This problem appeared in the magazine Mathematical Reflections, 2017 Issue 4 problem J424. Weird to see Iran use a geo problem from a known source. P.S. here is another solution Observe that AGDB is cyclic (1) and that problem statement is equivalent to proving \angle AFE = \angle GCB but because of (1) it is enough to prove that \triangle AEF \sim \triangle BGC or \frac{AE}{AF}=\frac{BC}{BG} (2). From the problem condition we get \frac {AE}{BF}=\frac{DE}{CF}=\frac{AE+DE}{BF+CF}=\frac{AD}{BC} (3). Also \triangle ADF \sim \triangle BGF hence \frac{AF}{BF}=\frac{AD}{BG} (4). Dividing (3) by (4) we get (2). (Note, aops thinks $ sign for latex is a picture and says "new users can not post pictures" that's why solution does not have latex, i will add latex code once i will be able to)
zbghj2
17.06.2021 02:53
Let circumcircle of CFG cut the extension of BG at point H. => FH is the diameter. => $FC\perp HC $ On AD, find a point K such that KF is tangent to the circumcircle of CFG. => $KF\perp HF $ => $\triangle DKF \sim \triangle CFH$ and $\triangle AKF \sim \triangle BFH$ => $\frac{AK} {BF} = \frac{KF} {FH} = \frac{DK} {CF} $ So K is E.
IMO2022Goldinshallah
26.10.2021 22:24
Oson ekanku bu masala. Shuni ham shuncha o'yladinglarmi. Very easy and nice problem.