Problem 1685

### Part (a)

1. Identify the given elements and their properties:
- $ABC$ is a right-angled triangle with $\angle C = 90^\circ$.
- $K$, $L$, $M$ are the midpoints of sides $AB$, $BC$, and $CA$ respectively.
- $N$ is a point on side $AB$.
- $CN$ meets $KM$ and $KL$ at points $P$ and $Q$ respectively.
- $S$ and $T$ lie on $AC$ and $BC$ respectively such that $APQS$ and $BPQT$ are cyclic quadrilaterals.

2. **Prove that $SQLC$ and $TPMC$ are squares:**
- Since $P$ lies on $KM$, $PC = PA$.
- $\angle APQ = \angle ACP + \angle CAP = 2 \cdot \angle ACP = 90^\circ$.
- Therefore, $SQ \perp AC$.
- Also, $QC$ bisects $\angle SCL$, making $SQLC$ a square.
- Similarly, $MPTC$ is a square.

3. **Show that $ML$, $ST$, and $CN$ concur:**
- Let $X = ML \cap ST$.
- Consider the reflection $r$ about $CN$.
- From the claim, $r$ maps $L$ to $S$ and $T$ to $M$.
- Thus, $r$ maps $ML$ to $TS$ and $TS$ to $ML$.
- Suppose $X'$ is the reflection of $X$ about $CN$.
- Therefore, $r$ takes $X = ML \cap TS$ to $X' = TS \cap ML = X$.
- Hence, $X = X' \Rightarrow X$ lies on $CN$.
- Therefore, $ML$, $ST$, and $CN$ concur.

$\blacksquare$

### Part (b)

1. Identify the given elements and their properties:
- $ABC$ is a right-angled triangle with $\angle C = 90^\circ$.
- $K$, $L$, $M$ are the midpoints of sides $AB$, $BC$, and $CA$ respectively.
- $N$ is a point on side $AB$.
- $CN$ is an altitude.
- $S$ and $T$ lie on $AC$ and $BC$ respectively such that $APQS$ and $BPQT$ are cyclic quadrilaterals.

2. **Prove that $PT \parallel NL$ and $QS \parallel NM$:**
- Since $Q$ lies on $KL$, $QC = QB \Rightarrow \angle CPT = \angle CBQ = \angle QCB = 90^\circ - \angle B$.
- As $L$ is the midpoint of the hypotenuse $BC$ of triangle $BNC$, $LN = LC$.
- Therefore, $\angle CNL = 90^\circ - \angle B$.

3. **Use coordinate geometry to show that $ST$ bisects $ML$:**
- Let $C = (0,0)$, $A = (0,a)$, $B = (b,0)$.
- Therefore, $L = \left(\frac{b}{2}, 0\right)$ and $M = \left(0, \frac{a}{2}\right)$.
- Since $CN \perp AB$, the slope of $CN$ is $\frac{b}{a}$.
- The equation of $CN$ is $y = \frac{b}{a}x$.
- The equation of line $AB$ is $y = \frac{-a}{b}x + a$.
- Equating the equations, we get $N = \left(\frac{a^2b}{a^2 + b^2}, \frac{ab^2}{a^2 + b^2}\right)$.
- The equation of $CN$ is $y = \frac{b}{a}x$.
- Therefore, $Q = \left(\frac{b}{2}, \frac{b^2}{2a}\right)$ and $P = \left(\frac{a^2}{2b}, \frac{a}{2}\right)$.
- Let $T = (x_0, 0)$.
- From the claim, $TP \parallel NL$.
- The slope of $NL$ is $\frac{\frac{b^2a}{a^2 + b^2}}{\frac{a^2b}{a^2 + b^2} - \frac{b}{2}} = \frac{2b^2a}{2a^2 - a^2b - b^3} = \frac{2ab}{a^2 - b^2}$.
- The slope of $PT$ is $\frac{\frac{a}{2}}{\frac{a^2}{2b} - x_0}$.
- Therefore, $x_0 = \frac{a^2 + b^2}{4b}$.
- Thus, $T = \left(\frac{a^2 + b^2}{4b}, 0\right)$ and $S = \left(0, \frac{a^2 + b^2}{4a}\right)$.
- Let $X = ML \cap ST$.
- The equations of lines $ML$ and $ST$ are $y = \frac{-a}{b}x + \frac{a}{2}$ and $y = \frac{-b}{a}x + \frac{a^2 + b^2}{4a}$ respectively.
- Equating, we get:
$$x = \frac{\frac{b^2 - a^2}{4a}}{\frac{b^2 - a^2}{ab}} = \frac{b}{4}$$
- Therefore, $y = \frac{a}{4}$, which gives $X$ as the midpoint of $ML$.

$\blacksquare$