Geometry (Finding angle $x$)

Question

So everybody was given this question:

in the recent new zealand mathematics exam yesterday (for students aged 14-15), and most if not everybody was stumped by it.

How would you solve this?

Answer 1

I think you mean that $B$ and $G$ they are mid-points of $FH$ and $AC$ respectively.

Since $FG=\frac{1}{2}BG= \frac{1}{2}DG,$ we obtain: $$\measuredangle DGB=\measuredangle FDG=30^{\circ}.$$ Thus, $$\measuredangle DBE=2\measuredangle DBG=2\cdot75^{\circ}=150^{\circ}.$$

Answer 2

There isn't enough information. If we assume $G$ is the midpoint of $FH$ or if we assume the kite is symmetrically aligned with the square so that line $DE$ is parallel with $FH$ then by Michael Rozenberg's and Deepak's deleted answer we can conclude that the angle is $150$.

But if the kite is inserted into the square at an angle so that $G$ is not the midpoint and $DE$ is not parallel, we can not conclude anything. Suppose we place point $B$ on line $AC$ slightly to the left of $G$ on line $FH$. There will be one pair of points $D,E$ so that $BG= DG= EF$, but those might not forma a kite if $BD \ne BE$. However but sliding $B,G$ to the left or right we will force point $D$ lower or higher and point $E$ higher or lower, and line $DB$ longer or shorter and line $BE$ shorter or longer. The will be one precise point where $DB = BE$ and an assymetric kite was placed in the square.

There are several such kites depending on how "askance" points $B$ and $G$ are from each other. But they will all form angles $DBE$ that are $\ge 150$, if I am not mistaken[$*$], with $BDE = 150$ precisely at the one symmetrical embedded kite.

Somehow, I suspect the original problem either indicated the kite was inserted symmetrically, or it was assumed.

[$*$] A big "if". But I'm pretty sure the wider "askew" the kite is aligned, the larger angle $DBE$ will be.