# Show three points are collinear

We have trapezoid $$ABCD$$ $$(AB||CD)$$. $$\angle BAD + \angle ABC = 90^\circ$$. $$M,N,P$$ and $$Q$$ are the midpoints of $$AB,CD,AC$$ and $$BD$$, respectively. $$AD$$ intersects $$BC$$ in $$K$$. Are $$M,N,K$$ collinear?

I tried to make the graph in GeoGebra, but I didn't succeed. I am sorry. $$\angle BAD+\angle ABC=90^\circ$$ is the same as $$\angle CDK+\angle DCK=90^\circ$$, thus $$\angle AKB=90^\circ$$. How to show that $$M$$ lies on $$KN$$?

$$P,Q$$ and the angles $$ABC,BAD$$ are irrelevant (except insofar as they guaraantee that the lines $$AD,BC$$ intercept).
Since $$AB$$ is parallel to $$DC$$ the triangles $$KAB,KDC$$ are similar. $$N$$ is the midpoint of $$DC$$ and $$M$$ is the midpoint of $$AB$$, so $$K,N,M$$ are collinear. [An expansion centre $$K$$, by a factor $$KA/KD$$ takes $$N$$ to $$M$$.]
• I suggest looking up SimilarTriangles on Wikipedia or in your textbook. Two triangles are similar if they have the same shape, but maybe different sizes. One may be a smaller version of the other. So $KDC,KAB$ similar means that $\angle DKC=\angle AKB$, and $\angle KDC=\angle KAB$ (and hence $\angle DCK=\angle ABK$. It is hard to do this problem without either similar triangles or something more advanced. Sep 29 '19 at 18:55