$A, B, C, D, E$ points are on $\Gamma$ cycle clockwise. $[AE \cap [CD = \{M\}$ and $[AB \cap [DC = \{N\}$. The line parallels to $EC$ and passes through $M$ intersects with the line parallels to $BC$ and passes through $N$ on $K$. Similarly, the line parallels to $ED$ and passes through $M$ intersects with the line parallels to $BD$ and passes through $N$ on $L$. Show that the lines $LD$ and $KC$ intersect on $\Gamma$.
Problem
Source: Turkey EGMO TST 2020 #5
Tags: geometry, geometry proposed
Prod55
15.07.2020 18:39
$\rm KM\parallel EC,KN\parallel BC\Rightarrow \angle NKM=\angle EBC$.Similarly $\rm \angle KNM=\angle BDE$.So we have that $\rm \angle NKM=\angle BCE=\angle BDE=\angle NLM=\angle NAM$.Therefore the points $\rm A,M,N,K,L$ are concyclic,let $\rm (w)\equiv (A,M,N,K,L),T\equiv (\Gamma)\cap (w)$.I wil show that the points $\rm K,T,C$ are collinear. Let $\rm I$ the inverse with pole $\rm C$ and power $\rm CM\cdot CN$.Set $\rm I(A)=A' \in (w),(D)=D'$.The points $\rm A,A',D,D'$ are concuclic so $\rm \angle D'A'A=\angle ADC=\angle DMA+\angle EAD=\angle DMA+\angle ECD\overset{KM\parallel EC}{=}\angle DMA+\angle KMN=180^{\circ}-\angle AMK=\angle KA'A$ therefore the points $\rm A',K,D'$ are collinear so $\rm I(K)\in (\Gamma)$ so $\rm I(K)\equiv T$.Therefore the points $\rm K,T,C$ are collinear.Similarly the points $\rm L,T,D$ are collinear ,the proof is complete.
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415122
15.07.2020 18:50
Which app did you use to make the figure?
Prod55
15.07.2020 19:05
Vivek1295_9 wrote: Which app did you use to make the figure? Hi,i use Geogebra.