In an acute triangle $ABC$ with $AC > AB$, let $D$ be the projection of $A$ on $BC$, and let $E$ and $F$ be the projections of $D$ on $AB$ and $AC$, respectively. Let $G$ be the intersection point of the lines $AD$ and $EF$. Let $H$ be the second intersection point of the line $AD$ and the circumcircle of triangle $ABC$. Prove that \[AG \cdot AH=AD^2\]
Problem
Source: 2013 Baltic Way, Problem 11
Tags: geometry, circumcircle, trigonometry, geometry unsolved
pi37
31.12.2013 05:10
Let $O$ be the circumcenter of $\triangle ABC$; let $AO$ intersect $BC$ at $J$ and the circumcircle of $ABC$ at $K$. Since $AH$ is isogonal to $AO$, $KH\parallel DJ$, so $\frac{AD}{AH}=\frac{AJ}{AK}$. But note that $AFE\sim ABC$, and $D$ is the antipode of $A$ with respect to the circumcircle of $AFE$, so $G,D$ correspond to $J,K$ in the similar triangles. Thus \[ \frac{AD}{AH}=\frac{AJ}{AK}=\frac{AG}{AD}\Rightarrow AG\cdot AH=AD^2 \] as desired.
shivangjindal
31.12.2013 10:42
Simply observe that, $\angle ADF = B, \angle AGF = 90+B-C$ and $ \angle ABG = 90+B-C , \angle AHB = C$. Now using sine rule in triangle $AFG,ABH$ we get $AG \cdot AH = AF \cdot AB = AD \sinh \cdot AD %Error. "cosecB" is a bad command. = AD^2 \Box$
ThirdTimeLucky
31.12.2013 14:45
$ AFDE $ is cyclic, so $ \angle BHA=\angle BCA=\angle ADF=\angle AEF $, so $ BEGH $ is cyclic and $ AG.AH=AE.AB=AD^2 $ as $ \angle ADE=\angle ABC $.
KRIS17
07.09.2019 19:58
Solution using only Similar Triangles: Let $K$ be the projection of $G$ onto line $DF$. So $GK$ is parallel to $AF$, so $\triangle DGK$ is similar to $\triangle DAF -->(1)$ Since $ABHC$ and $AEDF$ are cyclic, in $\triangle ABH$ and $\triangle GDF$, $\angle BAH$ = $\angle GFD$ and $\angle BHA$ = $\angle ACB$ = $\angle GDF$. Thus $\triangle ABH$ and $\triangle GDF$ are similar as well $ --> (2)$ From $(1)$: $FK/FD$ = $AG/AD$ From $(2)$: $FK/FD$ = $AD/AH$ Hence, $AG/AD$ = $AD/AH$, or $AG \cdot AH=AD^2$
rafaello
14.08.2021 00:18
Claim. $AD$ is tangent to $(BED)$ and $(CFD)$. Proof. Obvious due to right angles. Claim. $BEGH$ is cyclic. Proof. Note that by the first claim, $BEFC$ is cyclic, now $$\measuredangle BEG=\measuredangle FEA=\measuredangle BCA=\measuredangle BHG,$$the claim follows. Now, by PoP, $$AD^2=AE\cdot AB=AG\cdot AH,$$we are done.
Quidditch
01.10.2022 18:13
Perform an inversion center at $A$ with radius $AD$. Note that $E,F$ is send to $B,C$. So line $EF$ is send to $(ABC)$ and $AD$ is send to $AD$ itself. But then $AD\cap EF=G$ is send to $AD\cap (ABC)=H$ which gives $AG\cdot AH=AD^2$, as desired.
parthvartak
24.10.2022 17:48
We can see that $\triangle ADC \sim \triangle AFD \Rightarrow AF\times AC=AD^2$ Since $\angle AED +\angle AFD=90+90=180 \Rightarrow AEDF$ is cyclic. $\therefore \angle AFG=\angle AFE=\angle ADE=90-\angle DAE=\angle ABC=\angle AHC=\angle GHC\\ \therefore GHCF$ is cyclic. So, by Power of Point A, $AG\times AH=AF\times AC=AD^2!!!$ Hence we are done!