Let $E$ be an interior point of the convex quadrilateral $ABCD$. Construct triangles $\triangle ABF,\triangle BCG,\triangle CDH$ and $\triangle DAI$ on the outside of the quadrilateral such that the similarities $\triangle ABF\sim\triangle DCE,\triangle BCG\sim \triangle ADE,\triangle CDH\sim\triangle BAE$ and $ \triangle DAI\sim\triangle CBE$ hold. Let $P,Q,R$ and $S$ be the projections of $E$ on the lines $AB,BC,CD$ and $DA$, respectively. Prove that if the quadrilateral $PQRS$ is cyclic, then \[EF\cdot CD=EG\cdot DA=EH\cdot AB=EI\cdot BC.\]
Problem
Source: Baltic Way 2011
Tags: geometry proposed, geometry
skytin
06.11.2011 15:40
Hints : Let E' is isogonal to E wrt ABC E'BCD ~ AFEI , AFBE , IAED , EBGC , DECH are cycic
KRIS17
19.08.2019 01:01
Based on the given similarities of the triangles, it is easy to see that quadrilaterals $AFBE$ and $DECH$ are similar. Thus, we have: $AB/DC$ = $FE/EH$ or $AB.EH$ = $FE.DC$ Same way, since quadrilaterals $DIAE$ and $CEBG$ are similar, we get: $EG.DA$ = $EI.BC$ It remains to be proven that: $EG.DA$ = $AB.EH$ as the fourth eqiality follows directly form these 3 eualities. Since quadrilateral $PQRS$ is cyclic, we get that: $\angle RSE$ = $\angle RDE$ = $\angle BAF$ (from similarities given) and $\angle ESP$ = $\angle EAP$ Thus, we have $\angle RSP$ = $\angle EAF$. In the same manner we can derive that $\angle PQR$ = $\angle FBE$. Since, $\angle RSP$ and $\angle PQR$ are supplementry, so are $\angle EAF$ and $\angle FBE$. Thus, quadrilateral $AFBE$ is cyclic. Since quadrilaterals $AFBE$ and $DECH$ are similar, $DECH$ is cyclic too. In the same manner, we can derive that, quadrilaterals $DIAE$ and $CEBG$ are cyclic too. Applying Ptolemy's theorem to $AFBE$, we get: $AB.EF$ = $AF.EB$ + $FB.AE$ or $AB$ = $(AF/EF).EB + (FB/EF).AE$ = $(DE/EH).EB + (EC/EH).AE$ (from similarity of quadrilaterals $AFBE$ and $DECH$) or $AB.EH$ = $DE.EB$ + $EC.AE ---> (1)$ In a similar manner, by applying Ptolemy's theorem to $DIAE$, we get: $EG.AD$ = $DE.EB$ + $EC.AE ---> (2)$ From $(1)$ and $(2)$, we have $EG.DA$ = $AB.EH$ which completes the proof.
rafaello
14.08.2021 00:10
Claim. $ABEF$, $BECG$, $CEDH$ and $AEDI$ are cyclic. Proof. We have \begin{align*}\measuredangle IAE&=\measuredangle SAE+\measuredangle EBQ\\&=\measuredangle SPE+\measuredangle EPQ\\&=\measuredangle SPQ\\&=\measuredangle SRQ\\&=\measuredangle SRE+\measuredangle ERQ\\&=\measuredangle SDE+\measuredangle ECQ\\&=\measuredangle SDE+\measuredangle IDA\\&=\measuredangle IDE, \end{align*}we do the same for others, the claim follows. Claim. There exists $T$, the isogonal conjugate of $E$ wrt $ABCD$. Proof. By six-point circle theorem, we conclude that there exists $T$, the isogonal conjugate of $E$. Now, note that angle chasing implies that $\triangle GBE\sim\triangle CTD$ and $\triangle FBE\sim\triangle ATD$, thus $$\frac{EF}{EG}=\frac{AD}{DC}.$$Similarly, $\dfrac{EG}{EH}=\dfrac{AB}{AD}$, $\dfrac{EH}{EI}=\dfrac{BC}{AB}$ and $\dfrac{EI}{EF}=\dfrac{CD}{BC}$. This yields that $$EF\cdot CD=EG\cdot DA=EH\cdot AB=EI\cdot BC.$$