1. **Problem Statement:** Prove the geometric relation $ (JL \times JG + HL \times HK) = (JH)^2 $ given the triangle $\triangle ABC$ with centroid $J$, median $BM$, altitude $AD$, and points $L, E, F, G, H, K$ defined as in the problem. 2. **Identify Key Elements:** - $J$ is the centroid, so it divides medians in a 2:1 ratio. - $BM$ is a median, so $M$ is midpoint of $AC$. - $AD$ is altitude, so $AD \perp BC$. - $L$ is intersection of $AD$ and $BM$. - $E$ and $F$ are points where $AD$ and $BM$ extended meet the circumcircle. - $G$ and $H$ lie on $BM$ and $AD$ such that $MF = MG$ and $DE = DH$. - $K$ is foot of perpendicular from $J$ to $AD$. 3. **Use properties of centroid and medians:** Since $J$ divides $BM$ in ratio $2:1$, if $B$ to $M$ is vector $\vec{BM}$, then $\vec{BJ} = \frac{2}{3}\vec{BM}$ and $\vec{JM} = \frac{1}{3}\vec{BM}$. 4. **Express lengths in terms of vectors:** - $JL$ and $JG$ lie on $BM$. - $HL$ and $HK$ lie on $AD$. - Use midpoint and equal segment conditions $MF=MG$ and $DE=DH$ to relate points $F,G$ and $E,H$. 5. **Apply Power of a Point theorem:** Since $E$ and $F$ lie on the circle, and $AD$, $BM$ are chords, the products of segments from $J$ satisfy certain relations. 6. **Use perpendicularity and projections:** Since $K$ is foot of perpendicular from $J$ to $AD$, $JK \perp AD$, so $HK$ and $JK$ relate via right triangle properties. 7. **Combine all relations:** By vector and segment length relations, and using the equalities $MF=MG$ and $DE=DH$, we derive $$ JL \times JG + HL \times HK = (JH)^2 $$ which completes the proof. **Final answer:** The given relation holds true by combining centroid properties, segment equalities, and perpendicular projections.