Problem: The area of triangle $XYZ$ is 8 square inches. Points $A$ and $B$ are midpoints of congruent segments $\overline{XY}$ and $\overline{XZ}$ . Altitude $\overline{XC}$ bisects $\overline{YZ}$ . The area (in square inches) of the shaded region is
[asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.5,2)--(7.5,2)); draw((5,4)--(5,0)); fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey); label(scale(0.8)*"$X$", (5,4), N); label(scale(0.8)*"$Y$", (0,0), W); label(scale(0.8)*"$Z$", (10,0), E); label(scale(0.8)*"$A$", (2.5,2.2), W); label(scale(0.8)*"$B$", (7.5,2.2), E); label(scale(0.8)*"$C$", (5,0), S); fill((0,-.8)--(1,-.8)--(1,-.95)--cycle, white);[/asy]
$\textbf{(A)}\ 1\frac{1}2\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 2\frac{1}2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 3\frac{1}2$

Solution: $\usepackage[dvipsnames]{xcolor} \textcolor{BlueViolet}{\text{Super fast after convincing yourself they are congruent}}$ [asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.5,2)--(7.5,2)); draw((5,4)--(5,0)); fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey); label(scale(0.8)*"$X$", (5,4), N); label(scale(0.8)*"$Y$", (0,0), W); label(scale(0.8)*"$Z$", (10,0), E); label(scale(0.8)*"$A$", (2.5,2.2), W); label(scale(0.8)*"$B$", (7.5,2.2), E); label(scale(0.8)*"$C$", (5,0), S); fill((0,-.8)--(1,-.8)--(1,-.95)--cycle, white); /* Edit for solution */ /* A--C */ draw((2.5,2)--(5,0)); /* A--[half of YC] */ draw((2.5,2)--(2.5,0)); label(scale(0.8)*"$D$",(2.5,0),S); /* B--C */ draw((7.5,2)--(5,0)); /* B--[half of CZ] */ draw((7.5,2)--(7.5,0)); label(scale(0.8)*"$E$",(7.5,0),S); /* drawing right angles */ draw((2.5,0.5)--(2,0.5)); draw((2,0.5)--(2,0));  draw((2.5,0.5)--(3,0.5)); draw((3,0.5)--(3,0));  draw((7.5,0.5)--(7,0.5)); draw((7,0.5)--(7,0));  draw((7.5,0.5)--(8,0.5)); draw((8,0.5)--(8,0));  draw((5.5,2.5)--(4.5,2.5)--(4.5,1.5)--(5.5,1.5)--cycle);  label(scale(0.8)*"$F$", (5,2), NE);  label(scale(1)*"All credits for original unedited asymptote for the problem go to whoever made the asymptote in the 'Problem' section.", (5,-1), S); [/asy]
We draw the triangles as shown above. What seems like a half-square (or a corner of a square) indicates $90$ °. (This is not a random idea - the two original two triangles on the top inspires us to do so.)
Note that $\overline{AD}$ is perpendicular to $\overline{YC}$ $\overline{BE}$ is perpendicular to $\overline{CZ}$
$\overline{AB}$ is perpendicular to $\overline{YZ}$ , so because of corresponding angles theorem, $\angle YAD=\angle AXF$ . We are also told $\overline{YA}=\overline{AX}$ . Both $\bigtriangleup \text{XAF}$ and $\bigtriangleup \text{AYD}$ have a $90$ ° angle. Because of AAS Congruency (Angle-Angle-Side; $90$ °, $\angle YAD$ , and side $\overline{YA}$ ), $\bigtriangleup \text{XAF}$ is congruent to $\bigtriangleup \text{AYD}$ . Because of symmetry, the same goes for $\bigtriangleup \text{XFB}$ and $\bigtriangleup \text{BEZ}$
$\bigtriangleup \text{YAD}$ is congruent to $\bigtriangleup \text{CAD}$ , which is congruent to $\bigtriangleup \text{ACF}$ because of symmetry. So there are $4$ congruent triangles on the left side. There is also $4$ congruent triangles on the right side because of symmetry.
There are $8$ congruent triangles in all. We are told the area of the triangle $\bigtriangleup \text{XYZ}$ is 8. $8\div8=1$ , so each of the small congruent triangles is $1$ . The shaded area contains $3$ small triangles, so the area of the shaded section is $1\cdot3=\boxed{3}$

Answer: D 3