1. **Problem statement:** Given a right isosceles triangle $\triangle ABC$ with the right angle at $A$, $M$ is the midpoint of $BC$, and $E$ lies on segment $MC$. Lines $BH$ and $CK$ are drawn perpendicular to $AE$ with $H$ and $K$ on $AE$. We need to prove that $\triangle MHK$ is an isosceles right triangle. 2. **Setup and notation:** Since $\triangle ABC$ is right isosceles at $A$, assume $AB = AC = a$ and $\angle A = 90^\circ$. 3. **Coordinates assignment:** Place $A$ at the origin $(0,0)$, $B$ at $(a,0)$, and $C$ at $(0,a)$. 4. **Find midpoint $M$ of $BC$:** $$M = \left(\frac{a+0}{2}, \frac{0+a}{2}\right) = \left(\frac{a}{2}, \frac{a}{2}\right)$$ 5. **Point $E$ lies on segment $MC$:** Let $E$ divide $MC$ in ratio $t$ from $M$ to $C$: $$E = M + t(C - M) = \left(\frac{a}{2}, \frac{a}{2}\right) + t\left(0 - \frac{a}{2}, a - \frac{a}{2}\right) = \left(\frac{a}{2}(1 - t), \frac{a}{2}(1 + t)\right)$$ where $0 < t < 1$. 6. **Equation of line $AE$:** Since $A=(0,0)$ and $E=\left(\frac{a}{2}(1 - t), \frac{a}{2}(1 + t)\right)$, the vector $\overrightarrow{AE} = \left(\frac{a}{2}(1 - t), \frac{a}{2}(1 + t)\right)$. 7. **Find points $H$ and $K$:** - $H$ is the foot of perpendicular from $B=(a,0)$ to line $AE$. - $K$ is the foot of perpendicular from $C=(0,a)$ to line $AE$. Using projection formula for foot of perpendicular from point $P$ to line through $A$ in direction $\vec{v}$: $$H = A + \frac{\overrightarrow{AP} \cdot \overrightarrow{AE}}{\|\overrightarrow{AE}\|^2} \overrightarrow{AE}$$ Calculate $H$ for $P=B$: $$\overrightarrow{AB} = (a,0)$$ $$\overrightarrow{AE} = \left(\frac{a}{2}(1 - t), \frac{a}{2}(1 + t)\right)$$ $$\overrightarrow{AB} \cdot \overrightarrow{AE} = a \cdot \frac{a}{2}(1 - t) + 0 = \frac{a^2}{2}(1 - t)$$ $$\|\overrightarrow{AE}\|^2 = \left(\frac{a}{2}(1 - t)\right)^2 + \left(\frac{a}{2}(1 + t)\right)^2 = \frac{a^2}{4}[(1 - t)^2 + (1 + t)^2] = \frac{a^2}{4}[2 + 2t^2] = \frac{a^2}{2}(1 + t^2)$$ Therefore, $$H = \frac{\frac{a^2}{2}(1 - t)}{\frac{a^2}{2}(1 + t^2)} \overrightarrow{AE} = \frac{1 - t}{1 + t^2} \overrightarrow{AE} = \left(\frac{1 - t}{1 + t^2} \cdot \frac{a}{2}(1 - t), \frac{1 - t}{1 + t^2} \cdot \frac{a}{2}(1 + t)\right) = \left(\frac{a}{2} \frac{(1 - t)^2}{1 + t^2}, \frac{a}{2} \frac{(1 - t)(1 + t)}{1 + t^2}\right)$$ Similarly, for $K$ from $C=(0,a)$: $$\overrightarrow{AC} = (0,a)$$ $$\overrightarrow{AC} \cdot \overrightarrow{AE} = 0 + a \cdot \frac{a}{2}(1 + t) = \frac{a^2}{2}(1 + t)$$ So, $$K = \frac{\frac{a^2}{2}(1 + t)}{\frac{a^2}{2}(1 + t^2)} \overrightarrow{AE} = \frac{1 + t}{1 + t^2} \overrightarrow{AE} = \left(\frac{a}{2} \frac{(1 + t)(1 - t)}{1 + t^2}, \frac{a}{2} \frac{(1 + t)^2}{1 + t^2}\right)$$ 8. **Coordinates of $M$, $H$, and $K$ are:** $$M = \left(\frac{a}{2}, \frac{a}{2}\right)$$ $$H = \left(\frac{a}{2} \frac{(1 - t)^2}{1 + t^2}, \frac{a}{2} \frac{1 - t^2}{1 + t^2}\right)$$ $$K = \left(\frac{a}{2} \frac{1 - t^2}{1 + t^2}, \frac{a}{2} \frac{(1 + t)^2}{1 + t^2}\right)$$ 9. **Calculate vectors $\overrightarrow{MH}$ and $\overrightarrow{MK}$:** $$\overrightarrow{MH} = H - M = \left(\frac{a}{2} \frac{(1 - t)^2}{1 + t^2} - \frac{a}{2}, \frac{a}{2} \frac{1 - t^2}{1 + t^2} - \frac{a}{2}\right) = \frac{a}{2} \left(\frac{(1 - t)^2 - (1 + t^2)}{1 + t^2}, \frac{1 - t^2 - (1 + t^2)}{1 + t^2}\right)$$ Simplify numerator: $$(1 - t)^2 - (1 + t^2) = (1 - 2t + t^2) - (1 + t^2) = -2t$$ $$(1 - t^2) - (1 + t^2) = -2t^2$$ So, $$\overrightarrow{MH} = \frac{a}{2} \left(\frac{-2t}{1 + t^2}, \frac{-2t^2}{1 + t^2}\right) = \left(-\frac{a t}{1 + t^2}, -\frac{a t^2}{1 + t^2}\right)$$ Similarly, $$\overrightarrow{MK} = K - M = \frac{a}{2} \left(\frac{1 - t^2}{1 + t^2} - 1, \frac{(1 + t)^2}{1 + t^2} - 1\right)$$ Simplify numerator: $$\frac{1 - t^2 - (1 + t^2)}{1 + t^2} = \frac{-2 t^2}{1 + t^2}$$ $$\frac{(1 + t)^2 - (1 + t^2)}{1 + t^2} = \frac{1 + 2t + t^2 - 1 - t^2}{1 + t^2} = \frac{2t}{1 + t^2}$$ So, $$\overrightarrow{MK} = \frac{a}{2} \left(-\frac{2 t^2}{1 + t^2}, \frac{2 t}{1 + t^2}\right) = \left(-\frac{a t^2}{1 + t^2}, \frac{a t}{1 + t^2}\right)$$ 10. **Check if $\triangle MHK$ is right angled:** Calculate dot product: $$\overrightarrow{MH} \cdot \overrightarrow{MK} = \left(-\frac{a t}{1 + t^2}\right) \left(-\frac{a t^2}{1 + t^2}\right) + \left(-\frac{a t^2}{1 + t^2}\right) \left(\frac{a t}{1 + t^2}\right) = \frac{a^2 t^3}{(1 + t^2)^2} - \frac{a^2 t^3}{(1 + t^2)^2} = 0$$ So, $\overrightarrow{MH}$ is perpendicular to $\overrightarrow{MK}$, confirming a right angle at $M$. 11. **Check if $\triangle MHK$ is isosceles:** Calculate lengths: $$|\overrightarrow{MH}| = \sqrt{\left(-\frac{a t}{1 + t^2}\right)^2 + \left(-\frac{a t^2}{1 + t^2}\right)^2} = \frac{a}{1 + t^2} \sqrt{t^2 + t^4} = \frac{a t \sqrt{1 + t^2}}{1 + t^2}$$ $$|\overrightarrow{MK}| = \sqrt{\left(-\frac{a t^2}{1 + t^2}\right)^2 + \left(\frac{a t}{1 + t^2}\right)^2} = \frac{a}{1 + t^2} \sqrt{t^4 + t^2} = \frac{a t \sqrt{1 + t^2}}{1 + t^2}$$ Lengths are equal, so $\triangle MHK$ is isosceles. **Final conclusion:** $\triangle MHK$ is a right isosceles triangle with the right angle at $M$. **Answer:** $\boxed{\triangle MHK \text{ is an isosceles right triangle}}$