Prove that a particular angle is always 90°

Question

I'm stuck with this exercise since all we know is that DK = NC, meaning that the positions of K and N aren't fixed. I somehow figured that (KN) will always be parallel to (DC) and that KNA = NKB = 45° but that doesn't help me so much and I could'nt even explain why that's actually the case. help!

Answer 1

Since $\measuredangle KDC = \measuredangle NCB = 45^{\circ}$, $DC = CB$ (these $2$ come from $ABCD$ being a square with diagonals) and it's given that $DK = NC$, then $SAS$ gives that $\triangle KDC \cong \triangle NCB$. Thus, if we have

$$\measuredangle KCD = \measuredangle NBC = \theta \tag{1}\label{eq1A}$$

then we also have by the sum of angles in a triangle being $180^{\circ}$ in $\triangle CNB$ that

$$\measuredangle BNC = 180^{\circ} - 45^{\circ} - \theta = 135^{\circ} - \theta \tag{2}\label{eq2A}$$

By the sum of the angles along a straight line being $180^{\circ}$, we then also get

$$\measuredangle CNM = 180^{\circ} - (135^{\circ} - \theta) = 45^{\circ} + \theta \tag{3}\label{eq3A}$$

By subtracting angles and using \eqref{eq1A}, we next have

$$\measuredangle NCM = \measuredangle NCD - \measuredangle KCD = 45^{\circ} - \theta \tag{4}\label{eq4A}$$

By the sum of angles in a triangle being $180^{\circ}$ in $\triangle NCM$, we get that

$$\measuredangle NMC = 180^{\circ} - (45^{\circ} + \theta) - (45^{\circ} - \theta) = 90^{\circ} \tag{5}\label{eq5A}$$

Finally, using the sum of angles along a straight line being $180^{\circ}$ leads to

$$\measuredangle KMB = 180^{\circ} - 90^{\circ} = 90^{\circ} \tag{6}\label{eq6A}$$

---

Update: The above calculations can be simplified somewhat by using that the measure of any external angle of a triangle is equal to the sum of measures of the $2$ opposite internal angles. This allows us to bypass \eqref{eq2A} and replace \eqref{eq3A} with

$$\measuredangle CNM = \measuredangle NCB + \measuredangle NBC = 45^{\circ} + \theta \tag{7}\label{eq7A}$$

Similarly, \eqref{eq5A} could also be bypassed, with \eqref{eq3A} and \eqref{eq4A} being used to replace \eqref{eq6A} with

$$\measuredangle KMB = \measuredangle CNM + \measuredangle NCM = (45^{\circ} + \theta) + (45^{\circ} - \theta) = 90^{\circ} \tag{8}\label{eq8A}$$

Answer 2

If you rotate the entire figure $90^\circ$ clockwise, $BN$ moves exactly onto where $CK$ used to be. So those two lines must be orthogonal.

Answer 3

Argue as in @JohnOmielan's answer that $\measuredangle KCD=\measuredangle NBC$. Extend segment $BM$ to intersect the bottom segment of the square at (say) $P$. Looking at triangles $\triangle BCP$ and $\triangle CMP$, we find that they share angle $\angle CPM$, while $$\measuredangle PCM =\measuredangle KCD =\measuredangle NBC=\measuredangle PBC.$$ Conclude the remaining angles are equal: $\measuredangle CMP=\measuredangle BCP=90$. Since $\angle CMP$ and $\angle KMB$ are vertical angles, conclude $\angle KMB=90$.