I found this problem in a Geometry group on Facebook.
I know such problems may have errors but the OP claims it is correct.
We are given that $ABCD$ is a square.
Also $EG = GH = 4$ and $\angle EGH = 90^\circ$.
Also $GF = 10$.
We want to calculate the area of $DEGH$.
I tried to split it into two triangles, EGH and EDH.
Area of $EGH = \frac {1}{2}*4*4 = 8$.
Also $EH^2 = 4^2+4^2 = 32$.
Area of $EDH = \frac {1}{2}*DE*DH$ and $DE^2+DH^2 = 32$.
Rectangle DEGH is cyclic, so angles EDG and GDH are $45^\circ$.
Also, if we draw the altitudes of triangles GDH and GDE, these are equal (because a square is formed) and then the area of DEGH is equal to the area of this square.
However, I don't see how to calculate this altitude (or the side of the square).
I also see some similar right triangles with hypotenuses 4 and (10+4) but then I am missing the side of the triangle ABCD.
Any clues??
Thank you!
Best Answer
The desired area is not uniquely determined.
To see why, observe points $E, G, H, F$ are fixed. Then $D$ must lie on a semicircle with $EH$ as diameter. The foot of the perpendicular from $F$ to the line passing through $D$ and $H$ is the point $C$, which then determines the square $ABCD$. So we have fully constructed an entire family of admissible diagrams as a function of the variable point $D$, and since the area of $\triangle EDH$ is obviously not unique in such a semicircle, neither is the desired area.
In fact, we can make this construction explicit. Place the figure on a coordinate plane such that $$G = (0,0), \\ F = (10,0), \\ H = (0,4), \\ E = (-4,0).$$ Then the aforementioned semicircle has center at $O = (-2, 2)$ and $D$ is parametrized by some angle $\theta$ corresponding to the angle the radius $DO$ makes with the positive $x$-axis: $$D(\theta) = (-2 + 2 \sqrt{2} \cos \theta, 2 + 2 \sqrt{2} \sin \theta), \quad \pi/4 < \theta < 5\pi/4.$$
Then the line passing through $D$ and $H$ has equation
$$y - 4 = \frac{2 + 2 \sqrt{2} \sin \theta - 4}{-2 + 2 \sqrt{2} \cos \theta - 0} (x - 0),$$ or equivalently $$y = \frac{-1 + \sqrt{2} \sin \theta}{-1 + \sqrt{2} \cos \theta}x + 4.$$ The line through $F$ that is perpendicular to $DH$ is $$y - 0 = \frac{1 - \sqrt{2} \cos \theta}{-1 + \sqrt{2} \sin \theta}(x - 10).$$ The point of intersection of these two lines is $C$, which is $$C(\theta) = \left(5 - \frac{3}{\sqrt{2}} \cos \theta + \frac{7}{\sqrt{2}} \sin \theta, 2 - \frac{7}{\sqrt{2}} \cos \theta - \frac{3}{\sqrt{2}} \sin \theta \right).$$
It is a simple matter to find $A$ and $B$ from $C$ and $D$, but we don't need to show the calculation. We also do not need to show that the area of quadrilateral $DEGH$ is a function of $\theta$--this is obvious.
Finally, I have attached an animation showing the issue. We note that the lower bound of the angle $\theta$ is actually larger than $\pi/4$ because the segment $GF$ must lie inside the square; however, this does not change the fact that the desired area is not uniquely determined, since there is a continuous range of angles for $\theta$ that are valid. (As an exercise, what is this minimum $\theta$?) I trust this should completely settle the matter.