We have the following points on the Cartesian plane: $A(0,a),B(a,a),C(a,0),D(0,0),M(a,\frac a2)$. We know that $DM$ is on the line $y=\frac 12x$. $HC$ is on the line perpendicular to it, so it has a slope of $-2$. So we have $HC$ being on a line $y=-2x+b$, where $b$ is the y-intercept. We know $C$ is on this line, so plugging its coordinates, we get $y=-2x+b \implies 0=-2a+b \implies b=2a$ and the line is $y=-2x+2a$. Point H is the intersection of lines $DM$ and $HC$ so we have the system of equations $$y=\frac12x$$$$y=-2x+2a.$$Solving, we get $x=\frac45a,y=\frac25a$ so we have $H(\frac45a,\frac25a)$. The distance from $H$ to $A$ is $\sqrt{(\frac45a-0)^2+(\frac25a-a)^2}=\sqrt{\frac{16}{25}a^2+\frac{9}{25}a^2}=a$. Since $AB$ also has a length of $a$, we have $AB=AH$, as claimed.