Beautiful Geometry 1

Let the square be $ABCD$ and the rectangle be $AEFG$. Let $EF$ and $FG$ intersect $BD$ in points $J$ and $K$. We denote by $x$, $y$, $z$, $T$ the lengths of $DK$, $KJ$, $JB$, $BC$ respectively.

From $S(ABD)=S(AEFG)$, we get that $S(KDG)+S(BJE)=S(KJF)$, and since $△KDG$, $△BJE$, $△KJF$ are $90°-45°-45°$, we find that:

$x^2+y^2=z^2.$

Using the Pythagorean Theorem, we choose a point $L$ inside $△BCD$, such that $KL=x$ and $KL=y$, and $△KJL$ is right-angle.

We have

$∠DKL+∠LJB=360°-LKJ-KJL=270°,$

so using that $△JBL$ and $△DKL$ are isosceles, we get:

$∠BLD=90°+∠KLD+∠BLJ=270°-∠DKL-∠LJB=135°.$

Thus, the point $L$ lies on the circle with center $A$ and radius $AB$, and $AD=AL=AB$. Then, we see that $AK$ and $AJ$ are bisectors of $∠DAL$ and $∠LAB$ respectively. We conclude:

$∠KAJ=∠KAL+∠LAJ=\frac{∠DAL+∠LAB}{2}=45°.$