Given regular decagon $ABCDEFGHIJ$, and midpoint $M$ of $AB$, prove that $AD$, $CJ$, $EM$ concur

Question

Let there be a regular decagon $ABCDEFGHIJ$, and $M$ the midpoint of $AB$. Prove $AD$, $CJ$, and $EM$ are concurrent.

I was going to do this problem by making AD and CJ pass through a point K, and construct EK and EM, and prove EKM is a straight line, as shown in the diagram below. But I got stuck because I cannot prove that JC is parallel to AB, and AD is parallel to BC - this is the only step I need to prove in order to solve the question. Is there a quick way to do this? Any help is appreciated.

PS I also need to prove triangles KDC and AJK are congruent, but I think that this will be easier if I prove the parallel lines first.

Answer 1

Let the sides of the decagon be of unit length. Then: $$ EC=2\sin72°=4\sin36°\cos36°=4sc, $$ where I set for simplicity: $$ c=\cos36°={1+\sqrt5\over4},\quad s=\sin36°=\sqrt{1-c^2}. $$ If $K$ is the intersection of $AD$ and $JC$, we also have: $$ AK=CK=1,\quad DK=JK=JC-1=2c. $$ Let then $H$ be the projection of $K$ onto $AB$. We have: $$ MH=2c-(c+AM)=c-{1\over2}, \quad HK=s. $$ Note that $\angle ECJ=90°$ by Thales's theorem. We can then compute: $$ \tan(\angle JKM)=\tan(\angle HMK)={HK\over MH}={s\over c-1/2}, \quad \tan(\angle CKE)={EC\over CK}=4sc $$ and it is easy to check that these expressions are equal, because: $$ {1\over c-1/2}={4\over\sqrt5-1}=\sqrt5+1=4c. $$ Hence $\angle JKM=\angle CKE$ and as a cosequence $\angle AKM=\angle EKD$ and $MKE$ are aligned, as it was to be proved.

Answer 2

Let $AD$ intersect $JC$ at $X$. Let $O$ be the center of the dodecagon. Due to symmetry, $B, X, O$ are collinear. Easy to see that $ABX$ and $JXO$ are similar isosceles triangles. We have $EJ=2JO=2JX$ and $AM=\frac 12 AB=\frac 12 AX$, so $$\frac{EJ}{JX}=2=\frac{XA}{AM}.$$ Since $\angle MAX=\angle XJE$, it follows by SAS that $\triangle MAX\sim\triangle XJE$. Now it's easy to see that $E,X,M$ are collinear.