Ratio of lengths $QD/QC$ where $Q$ belongs to the side $CD$ of a square

Question

Let ABCD be a square, M the middle of (AB), N the middle of (AD), {P} = CN ∩ DM and {Q} = AP ∩ CD. Calculate $QD/QC$.

You can see down my idea and the drawing::

Okey! So first of all i thought of the Thales Theoreme but nope. Then, what about similarity of triangles. I could get that as the final result. DP is perpendicular on NC and in the traingle DPC one of the catet is half of the other. What can i do next? I thought of showing that Q is the half of DC by congruence. Any idea is welcome. Hopw you can heĺp me! Thank you!

Answer 1

Sketch of a solution by analytic geometry :

Take coordinates with center in $D$. Let the sidelength of the square be 2 units : therefore $C(2,0)$ and $A(0,2)$.

$$\begin{cases}\text{equ. of line CN : }& y&=&-\tfrac12x+1\\ \text{equ. of line DM : }& y&=&2x\end{cases}\tag{1}$$

Their intersection point is obtained by solving system (1). It is

$$P(x=2/5,y=4/5).$$

Besides, as line $AP$ has equation $y=ax+2$, and $P$ belongs to this line, one must have : $4/5=a 2/5 +2$ giving $a=-3$. As a consequence, the equation of line $AP$ is

$$y=-3x+2\tag{2}$$

Its intersection with axis $DC$ is obtained by taking $y=0$ in (2) giving the abscissa of $Q$ : $x_Q=2/3.$

Therefore the answer to your question is

$$QD/QC=(2/3)/(4/3)=1/2$$

Answer 2

Since $DM\perp CN$, then $PNAM$ is a cyclic quadrilateral. And since $AM=AN$, then$$\angle APN=\angle MPA$$and respective vertical angles$$QPC=\angle DPQ$$ Therefore, by the angle bisector theorem$$\frac{QD}{QC}=\frac{PD}{PC}$$But by similar triangles$$\frac{PD}{PC}=\frac{DN}{DC}=\frac{1}{2}$$Therefore$$\frac{QD}{QC}=\frac{1}{2}$$