This is also Kosovo TST 2019 P4

parmenides51 wrote: Given a rectangle $ABCD$ such that $AB = b > 2a = BC$, let $E$ be the midpoint of $AD$. On a line parallel to $AB$ through point $E$, a point $G$ is chosen such that the area of $GCE$ is $$(GCE)= \frac12 \left(\frac{a^3}{b}+ab\right)$$Point $H$ is the foot of the perpendicular from $E$ to $GD$ and a point $I$ is taken on the diagonal $AC$ such that the triangles $ACE$ and $AEI$ are similar. The lines $BH$ and $IE$ intersect at $K$ and the lines $CA$ and $EH$ intersect at $J$. Prove that $KJ \perp AB$. Can anybody tell me how to solve this please.

ErijonHasi1 wrote: parmenides51 wrote: Given a rectangle $ABCD$ such that $AB = b > 2a = BC$, let $E$ be the midpoint of $AD$. On a line parallel to $AB$ through point $E$, a point $G$ is chosen such that the area of $GCE$ is $$(GCE)= \frac12 \left(\frac{a^3}{b}+ab\right)$$Point $H$ is the foot of the perpendicular from $E$ to $GD$ and a point $I$ is taken on the diagonal $AC$ such that the triangles $ACE$ and $AEI$ are similar. The lines $BH$ and $IE$ intersect at $K$ and the lines $CA$ and $EH$ intersect at $J$. Prove that $KJ \perp AB$. Can anybody tell me how to solve this please. Show that $EHCG$ and $XYGH$ are cyclic. Just see what the area condition implies and you'll be able to see why those quadrilaterals are cyclic.

Because $(GCE)=\frac{1}{2}\cdot a\cdot \frac{a^2+b^2}{b},$ we get $EG=\frac{a^2+b^2}{b}=\frac{EC^2}{EF},$ and therefore $\angle ECG=90^\circ$ and $EHCG$ is cyclic with $EG$ being a diameter of the corresponding circle. But $DE\perp EG,$ and thus $DEA$ is tangent to this circle, and by construction of the location of $I,$ we have that $I$ is also on this circle (because $AE^2=AI\cdot AC$). So is point $B,$ by symmetry (wrt to $C$ and the diameter $EG$); now, $\angle IKH=(\overarc {BI}-\overarc {EH})/2=(\overarc {HC}-\overarc {IE})/2=\angle HJI,$ and therefore $KJIH$ is cyclic. Finally, $\angle KJH=\angle KIH=\angle EGH=\angle HED=\angle JEA,$ and therefore $KJ\parallel DA$ and $KJ\perp AB.$

tk1 wrote: Because $(GCE)=\frac{1}{2}\cdot a\cdot \frac{a^2+b^2}{b},$ we get $EG=\frac{a^2+b^2}{b}=\frac{EC^2}{EF},$ and therefore $\angle ECG=90^\circ$ and $EHCG$ is cyclic with $EG$ being a diameter of the corresponding circle. But $DE\perp EG,$ and thus $DEA$ is tangent to this circle, and by construction of the location of $I,$ we have that $I$ is also on this circle (because $AE^2=AI\cdot AC$). So is point $B,$ by symmetry (wrt to $C$ and the diameter $EG$); now, $\angle IKH=(\overarc {BI}-\overarc {EH})/2=(\overarc {HC}-\overarc {IE})/2=\angle HJI,$ and therefore $KJIH$ is cyclic. Finally, $\angle KJH=\angle KIH=\angle EGH=\angle HED=\angle JEA,$ and therefore $KJ\parallel DA$ and $KJ\perp AB.$ Leartia wrote: ErijonHasi1 wrote: parmenides51 wrote: Given a rectangle $ABCD$ such that $AB = b > 2a = BC$, let $E$ be the midpoint of $AD$. On a line parallel to $AB$ through point $E$, a point $G$ is chosen such that the area of $GCE$ is $$(GCE)= \frac12 \left(\frac{a^3}{b}+ab\right)$$Point $H$ is the foot of the perpendicular from $E$ to $GD$ and a point $I$ is taken on the diagonal $AC$ such that the triangles $ACE$ and $AEI$ are similar. The lines $BH$ and $IE$ intersect at $K$ and the lines $CA$ and $EH$ intersect at $J$. Prove that $KJ \perp AB$. Can anybody tell me how to solve this please. Show that $EHCG$ and $XYGH$ are cyclic. Just see what the area condition implies and you'll be able to see why those quadrilaterals are cyclic. Thanks.