XLIII OM - I - Problem 2

In square $ABCD$ with side length $1$, point $E$ lies on side $BC$, point $F$ lies on side $CD$, the measures of angles $EAB$ and $EAF$ are $20^{\circ}$ and $45^{\circ}$, respectively. Calculate the height of triangle $AEF$ drawn from vertex $A$.
The measure of angle $ FAD $ is $ 90^\circ - (20^\circ + 45^\circ) = 25^\circ $. From point $ A $, we draw a ray forming angles of $ 20^\circ $ and $ 25^\circ $ with rays $ AE $ and $ AF $, respectively, and we place a segment $ AG $ of length $ 1 $ on it (figure 2).
From the equality $ |AG| =|AB| = 1 $, $ | \measuredangle EAG| = | \measuredangle EAB| = 20^\circ $, it follows that triangle $ EAG $ is congruent to $ EAB $.
Similarly, from the equality $ |AG| = |AD| = 1 $, $ | \measuredangle FAG| = | \measuredangle FAD| = 25^\circ $, it follows that triangle $ FAG $ is congruent to $ FAD $.
This means that

point $ G $ lies on segment $ EF $ and is the foot of the altitude of triangle $ AEF $ drawn from vertex $ A $. Its length $ |AG| = 1 $.