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The area of triangle $ABC$ is double the area of $AED$.
But, $ADE$ and $ACB$ are similar triangles because $DE$ is parallel to $CB$$X$ is any suitable point on $AD$
$ZX$ is perpendicular to $AC$, and $ZX$ is equal in length to $AX$.
So $AXZ$ is an isosceles right-angled triangle.
By sweeping an arc centre $A$ from $X$ to $AZ$ at $N$, $AN$ is made equal to $AX$
$AN$ to $AZ$ is now in the required ratio.
Drawing from $N$ parallel to $ZC$ the point $D$ is reached.
Because $AND$ and $AZC$ are similar triangles, $AD$ and $AC$ are in the required ratio.
Excellent and simple!