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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 $KN$?

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$P,Q$ 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 $M$.]

enter image description here

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  • $\begingroup$ Thank you for the response! I haven't studied similar triangles. Can we show this in a different way? $\endgroup$
    – Stellar
    Sep 29 '19 at 18:21
  • $\begingroup$ What have you studied? $\endgroup$
    – almagest
    Sep 29 '19 at 18:27
  • $\begingroup$ This question is too general. $\endgroup$
    – Stellar
    Sep 29 '19 at 18:28
  • $\begingroup$ Similar triangles are one of the first things you learn in Euclidean geometry. If you don't know anything about them, the question is whether you know any Euclidean geometry at all. $\endgroup$
    – almagest
    Sep 29 '19 at 18:31
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    $\begingroup$ 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. $\endgroup$
    – almagest
    Sep 29 '19 at 18:55

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