Let $ABCD$ be a quadrilateral inscribed in circle of center $O$. The perpendicular on the midpoint $E$ of side $BC$ intersects line $AB$ at point $Z$. The circumscribed circle of the triangle $CEZ$, intersects the side $AB$ for the second time at point $H$ and line $CD$ at point $G$ different than $D$. Line $EG$ intersects line $AD$ at point $K$ and line $CH$ at point $L$. Prove that the points $A,H,L,K$ are concyclic, e.g. lie on the same circle.
Problem
Source:
Tags: geometry, Concyclic, cyclic quadrilateral, circumcircle
Gouss
26.10.2020 19:34
What is E?
parmenides51
26.10.2020 20:45
E is the midpoint of BC, just added it
songhongyi
11.01.2024 15:10
We only need to prove that $\angle A + \angle ELC = 180^{\circ}$. We have $\angle ZHC = \angle ZEC$ in $(ZHEC)$. Since $\angle ELC = \angle EHL + \angle HEL$, where $\angle EHL = \angle EZC = 90^{\circ}-\angle B$ and $\angle HEL = \angle HCG = \angle DCB - (90^{\circ}-\angle B)$, $\angle ELC = \angle DCB$. Since $\angle A + \angle DCB = 180^{\circ}$ in $(ABCD)$, proof is done.
Attachments:
parmenides51
30.04.2024 08:53
Let <ZCG=a, <ZCB=b We shalll prove that <ZAK=a+b=<HLK, threfore AKLH is cyclic <ZEC=<ZCG=a from cyclic ZGCH <ZBC=ZCB=b since ZBC is Isosceles with base BC Let O be the intersection of ZE and HC. <EOL=<HOZ=<HBE=b from cyclic BHOE (as <ZHC=90^o=ZEC from cyclic ZHECG) <HLG= <OEL+ <EOL=a+b (from triangle OLE) finally <ZAD=<GCB=b+a from cyclic ABCD Combining last two relations we get <ZAK=a+b=<HLK Since I cannot upload a photo by mobile, here is the related handwritten photo https://drive.google.com/file/d/1nG1pRECT0x3idvLUUVHF7x9xGHINWmgu/view?usp=drivesdk