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Given a trapezoid $ABDC$, and line segment $PQ$ where $P$ and $Q$ are points on $AC$ and $BD$, respectively, s.t. $AB||PQ||CD$. Suppose $PQ$ intersects $BC$ at $K$ and $AD$ at $J$, prove that $PK$ and $JQ$ are equal.

Below is an extreme case where $K$ and $J$ are at point $O$. Part of the reason must be because the lines are parallel. Also, I checked on GeoGebra and seen that they are numerically equal,i.e. $PK=JQ$.

enter image description here

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    $\begingroup$ $P$, $K$ and $J$ are collinear. How can $PK=JP$ if $J\ne K$? $\endgroup$ – CY Aries Apr 2 '18 at 15:35
  • $\begingroup$ Sorry for the confusion, I edited the question. $\endgroup$ – John Glenn Apr 2 '18 at 15:36
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As $\triangle ABC\sim\triangle PKC$, $AB:PK=BC:KC$.

As $\triangle ABD\sim\triangle JQD$, $AB:JQ=BD:QD$.

As $\triangle BCD\sim\triangle BKQ$, $BC:BK=BD:BQ$.

So, $BC:KC=BC:(BC-BK)=BD:(BD-BQ)=BD:QD$

Therefore, $AB:PK=AB:JQ$ and hence $PK=JQ$

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