1. **Problem statement:** Given square ABCD with M, N as midpoints of AB and BC respectively, and E as the intersection of CM and DN. We need to prove: a) CM is perpendicular to DN at E. b) With K as midpoint of DC and H as the foot of the altitude from A in triangle ADE, prove A, H, K are collinear. c) Prove AE = AB. 2. **Setup coordinates:** Let ABCD be a square with side length $s$. Place A at $(0,0)$, B at $(s,0)$, C at $(s,s)$, D at $(0,s)$. Then: - $M$ midpoint of $AB$ is $\left(\frac{s}{2},0\right)$ - $N$ midpoint of $BC$ is $\left(s,\frac{s}{2}\right)$ 3. **Find equations of lines CM and DN:** - $C=(s,s)$, $M=\left(\frac{s}{2},0\right)$ Slope of $CM$: $$m_{CM} = \frac{0 - s}{\frac{s}{2} - s} = \frac{-s}{-\frac{s}{2}} = 2$$ Equation of $CM$: $$y - s = 2(x - s) \Rightarrow y = 2x - s$$ - $D=(0,s)$, $N=\left(s,\frac{s}{2}\right)$ Slope of $DN$: $$m_{DN} = \frac{\frac{s}{2} - s}{s - 0} = \frac{-\frac{s}{2}}{s} = -\frac{1}{2}$$ Equation of $DN$: $$y - s = -\frac{1}{2}(x - 0) \Rightarrow y = -\frac{1}{2}x + s$$ 4. **Find intersection E of CM and DN:** Set: $$2x - s = -\frac{1}{2}x + s$$ Multiply both sides by 2: $$4x - 2s = -x + 2s$$ Bring terms together: $$4x + x = 2s + 2s \Rightarrow 5x = 4s \Rightarrow x = \frac{4s}{5}$$ Find $y$: $$y = 2x - s = 2 \times \frac{4s}{5} - s = \frac{8s}{5} - s = \frac{3s}{5}$$ So, $$E = \left(\frac{4s}{5}, \frac{3s}{5}\right)$$ 5. **a) Prove $CM \perp DN$ at $E$:** Slopes: $$m_{CM} = 2, \quad m_{DN} = -\frac{1}{2}$$ Since $2 \times \left(-\frac{1}{2}\right) = -1$, lines are perpendicular. They intersect at $E$, so $CM \perp DN$ at $E$. 6. **b) Prove points A, H, K are collinear:** - $K$ is midpoint of $DC$: $$D=(0,s), C=(s,s) \Rightarrow K=\left(\frac{s}{2}, s\right)$$ - $H$ is foot of altitude from $A$ to $DE$ in triangle $ADE$. Find equation of $DE$: $D=(0,s)$, $E=\left(\frac{4s}{5}, \frac{3s}{5}\right)$ Slope of $DE$: $$m_{DE} = \frac{\frac{3s}{5} - s}{\frac{4s}{5} - 0} = \frac{-\frac{2s}{5}}{\frac{4s}{5}} = -\frac{1}{2}$$ Equation of $DE$: $$y - s = -\frac{1}{2}(x - 0) \Rightarrow y = -\frac{1}{2}x + s$$ 7. **Find foot $H$ of perpendicular from $A=(0,0)$ to $DE$:** Slope of $DE$ is $-\frac{1}{2}$, so slope of perpendicular is $2$. Equation of perpendicular from $A$: $$y = 2x$$ Find intersection $H$ of: $$y = 2x$$ $$y = -\frac{1}{2}x + s$$ Set equal: $$2x = -\frac{1}{2}x + s \Rightarrow 2x + \frac{1}{2}x = s \Rightarrow \frac{5}{2}x = s \Rightarrow x = \frac{2s}{5}$$ Then: $$y = 2 \times \frac{2s}{5} = \frac{4s}{5}$$ So, $$H = \left(\frac{2s}{5}, \frac{4s}{5}\right)$$ 8. **Check collinearity of A, H, K:** Points: $$A=(0,0), H=\left(\frac{2s}{5}, \frac{4s}{5}\right), K=\left(\frac{s}{2}, s\right)$$ Calculate slopes: Slope $AH$: $$m_{AH} = \frac{\frac{4s}{5} - 0}{\frac{2s}{5} - 0} = \frac{4s/5}{2s/5} = 2$$ Slope $AK$: $$m_{AK} = \frac{s - 0}{\frac{s}{2} - 0} = \frac{s}{s/2} = 2$$ Since $m_{AH} = m_{AK}$, points $A, H, K$ are collinear. 9. **c) Prove $AE = AB$:** - $A=(0,0)$, $E=\left(\frac{4s}{5}, \frac{3s}{5}\right)$ Distance $AE$: $$AE = \sqrt{\left(\frac{4s}{5} - 0\right)^2 + \left(\frac{3s}{5} - 0\right)^2} = \sqrt{\frac{16s^2}{25} + \frac{9s^2}{25}} = \sqrt{\frac{25s^2}{25}} = s$$ - $AB$ is side length of square, so $AB = s$. Therefore, $$AE = AB$$ **Final answers:** a) $CM \perp DN$ at $E$. b) Points $A, H, K$ are collinear. c) $AE = AB$.