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

  • 1
    $\begingroup$ $P$, $K$ 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

As $\triangle ABC\sim\triangle PKC$, $AB:PK=BC:KC$.

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

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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.