Question Number 223905 by TonyCWX last updated on 09/Aug/25
$${ABCD}\:{is}\:{a}\:{square} \\ $$$${EL}={LF} \\ $$$${FN}={ND} \\ $$$${O}\:{is}\:{the}\:{center}\:{of}\:{square} \\ $$$${Prove}\:{that}\:{points}\:{K},\:{L},\:{O},\:{N}\:{and}\:{C}\:{are}\:{concyclic} \\ $$
Commented by TonyCWX last updated on 09/Aug/25
Commented by yerlow3 last updated on 09/Aug/25
Commented by Frix last updated on 09/Aug/25
$$\mathrm{I}\:\mathrm{did}\:\mathrm{not}\:\mathrm{see}\:{BE},\:{BF}\:\mathrm{marked}\:\mathrm{green}\:\mathrm{at}\:\mathrm{first}, \\ $$$$\mathrm{anyway}\:\mathrm{I}\:\mathrm{got}\:\mathrm{this}\rceil \\ $$$$\mathrm{Let}\:{O}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\wedge\mathrm{side}\:\mathrm{of}\:\mathrm{square}\:=\mathrm{2}, \\ $$$$\mathrm{further}\:{E}=\begin{pmatrix}{−\mathrm{1}}\\{{q}}\end{pmatrix}\:\mathrm{and}\:{F}=\begin{pmatrix}{{p}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$\Rightarrow \\ $$$${L}=\begin{pmatrix}{\frac{{p}−\mathrm{1}}{\mathrm{2}}}\\{\frac{{q}+\mathrm{1}}{\mathrm{2}}}\end{pmatrix}\:\:\:\:\:{N}=\begin{pmatrix}{\frac{{p}+\mathrm{1}}{\mathrm{2}}}\\{\mathrm{0}}\end{pmatrix} \\ $$$${K}=\begin{pmatrix}{−\frac{\mathrm{5}{p}+{pq}−{q}−\mathrm{1}}{{p}+{pq}−{q}−\mathrm{5}}}\\{−\frac{{p}+{pq}+\mathrm{3}{q}+\mathrm{7}}{{p}+{pq}−{q}−\mathrm{5}}}\end{pmatrix} \\ $$$$\Rightarrow \\ $$$$\mathrm{The}\:\mathrm{conic}\:{CKLON}\:\mathrm{has}\:\mathrm{the}\:\mathrm{formula} \\ $$$${x}^{\mathrm{2}} −\frac{{p}−\mathrm{1}}{{q}+\mathrm{1}}{y}^{\mathrm{2}} −\frac{{p}+\mathrm{1}}{\mathrm{2}}{x}+\frac{\left({p}−\mathrm{1}\right)\left({p}+\mathrm{1}\right)}{\mathrm{2}\left({q}+\mathrm{1}\right)}{y}=\mathrm{0} \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{for} \\ $$$$−\frac{{p}−\mathrm{1}}{{q}+\mathrm{1}}=\mathrm{1}\:\Leftrightarrow\:{q}=−{p}\:\Leftrightarrow\:\mid{BE}\mid=\mid{BF}\mid \\ $$
Commented by TonyCWX last updated on 09/Aug/25
$$\:\underline{\underbrace{\lesseqgtr}} \\ $$
Answered by yerlow3 last updated on 09/Aug/25
$${LN}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{ED}\:\left(\mathrm{Using}\:\mathrm{the}\:\mathrm{midsegment}\:\mathrm{theorem}.\right. \\ $$$$\mathrm{LN}\:\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{ED},\:\mathrm{So}: \\ $$$$\angle{LON}\:=\:\angle{LCN} \\ $$$$\mathrm{Using}\:\mathrm{Angle}\:\mathrm{Equalities}\:\mathrm{that}\:\mathrm{transfee}\:\mathrm{to}\:\mathrm{point}\:{K}, \\ $$$$\angle{LKN}\:=\:\angle{LCN} \\ $$$$\mathrm{So}, \\ $$$$\mathrm{Points}\:\mathrm{K},\mathrm{L},\mathrm{O},\mathrm{N},\mathrm{C}\:\mathrm{Are}\:\mathrm{Concyclic}\checkmark \\ $$
Commented by TonyCWX last updated on 09/Aug/25
$${Are}\:{you}\:{sure}\:\angle{LON}\:=\:\angle{LCN}? \\ $$
Commented by yerlow3 last updated on 09/Aug/25
$$\mathrm{Because}\:\mathrm{if}\:\mathrm{Points}\:{K},{L},{O},{N},{C}\:\mathrm{lie}\:\mathrm{on}\:\mathrm{the}\:\mathrm{same}\:\mathrm{circle} \\ $$$$\mathrm{Line}\:\mathrm{Segment}\:{LN}\:\mathrm{is}\:\mathrm{also}\:\mathrm{chord}\:\mathrm{that} \\ $$$$\mathrm{Subtends}\:\mathrm{angles}\:\mathrm{at}\:\mathrm{O}\:\mathrm{and}\:\mathrm{C}.\:\mathrm{Hence}, \\ $$$$\angle{LON}\:=\:\angle{LCN}.\:\mathrm{So}\:\mathrm{im}\:\mathrm{sure} \\ $$
Commented by TonyCWX last updated on 09/Aug/25
$${Well},\:\angle{LON}\:{is}\:{actually}\:\mathrm{180}−\angle{LCN} \\ $$