The bisector of angle $BAD$ in parallelogram $ABCD$ intersects the lines $BC$ and $CD$ at the points $K$ and $L$ respectively. Prove that the center of the circle passing through the points $C$, $K$, and $L$ lies on the circle passing through the points $B$, $C$, and $D$.

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I noticed there're three isosceles triangles that I could think of: $\triangle{ABK}$, $\triangle{KCL}$ and $\triangle{ADL}$.

Now, to show that they're similar: $\angle{ABK}=\angle{ADL}$ as $ABCD$ is a parallelogram, furthermore, since $KC||AD$, $\angle{ADL}=\angle{KCL}$. Also, $\angle{BAK}=\angle{LAD}$ as $AK$ is angle bisector.

Since $KC||AD$, $\angle{LKC}=\angle{LAD}$. Thus, $\triangle{ABK}\sim\triangle{LDA}\sim\triangle{LCK}$ by AA similarity.

But I still haven't proven that the triangles are indeed isosceles and where do I go from here?

  • $\begingroup$ I also notice $\angle{BKA}=\angle{CKL}$ as they're verticle angles. $\endgroup$ – user683941 Jun 22 '19 at 2:07
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    $\begingroup$ You can fairly easily show those $3$ triangles are isosceles using $\angle{BAD} = \angle{BCD}$, $\angle{BAD}$ is supplementary to $\angle{ADC}$, $\angle{ADC} = \angle{ABC}$, and the sums of angles in a triangle is $180$ degrees (or $\pi$ radians if you prefer to work with radians). You can then show they're all similar based on your determined angles. $\endgroup$ – John Omielan Jun 22 '19 at 2:14

Let $E$ be a center of the circle $KCL$, $F$ be an intersection points of $AD$ with the circle $BDC$,

which different from $D$ and $\measuredangle BAK=\alpha$.

Thus, $$\measuredangle FBC=\measuredangle BCD=2\alpha$$ and since $$BK=BA=CD=BF,$$ we obtain $$\measuredangle BKF=90^{\circ}-\alpha.$$

In another hand, $$\measuredangle KEC=2\measuredangle KLC=2\alpha,$$ which gives $$\measuredangle EKC=90^{\circ}-\alpha,$$ which says points $E$, $K$ and $F$ are placed on the same line $FE$.

But $$\measuredangle FBC=\measuredangle FEC=2\alpha,$$ which says that $E$ is placed on the circle $BDC$ and we are done!

  • $\begingroup$ Oops, I just edited the post and changed the diagram cause I was working on it. Didn't notice you posted an answer $\endgroup$ – redsunx Jun 22 '19 at 6:57
  • $\begingroup$ @MC Urist It's OK. I added previous points in my proof. :) Thank you that you restored it. $\endgroup$ – Michael Rozenberg Jun 22 '19 at 7:03
  • $\begingroup$ I thought I had things worked out, but couldn't prove the first two equations. Somehow it didn't occur to me BC and FD were parallel... $\endgroup$ – redsunx Jun 22 '19 at 7:12
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    $\begingroup$ @MC Urist I think, it's a very nice problem. $\endgroup$ – Michael Rozenberg Jun 22 '19 at 7:14
  • $\begingroup$ Can this be proven using isosceles triangles, congruency or similarity? The hint says to show that there must be a few triangles that are congruent, so that's what I was trying to do in my original post. $\endgroup$ – user683941 Jun 23 '19 at 9:20

(1) Circle (Q) circumscribes $\triangle BCD $ and Circle (O) circumscribes $\triangle LCK$ but the exact location of O is not known at this stage.

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(2) LKA, the angle bisector of A, can be translated to TC. When produced, TC will cut AB and circle (Q) at R and S respectively. Note that AKCR is a parallelogram.

(3) Let CC’ be the common chord. Since $\triangle LAB$ is isosceles, we have $\angle 1 = \angle 2 =\angle 3 =\angle 4$, and therefore TC is tangent to circle (O) at C.

(4) Let the external angle bisector of $\angle BCD$ cut circle (Q) at P. By the internal and external angle bisector relationship, $\angle TCP = \angle PCS = 90^0$. This means O is somewhere on the line CP extended. Let say it is at $O_1$.

(5) By “the converse of angle in semi-circle”, $\angle PCS = 90^0$ implies PQS is the diameter of circle (Q) and therefore PQS is a straight line.

(6) BY noting that SCPC’ is cyclic with $\angle SCP = \angle SC’P = 90^0$, we can say SC’ is another tangent to the circle KCL at C’. Hence, O is somewhere on C'P extended. Let say it is at $O_3$.

(7) P, which is a point on the circle (Q), is the only candidate that meets the conditions found in (4) and (6). Therefore, O = P.


Denote by $P(X, o)$ the power of a point $X$ with respect to the circle $o$. Let $O$ be the center of circumcircle $o$ of triangle $KCL$ and let $r$ be the radius of that circle. Because $BK||AD$ we have $\sphericalangle BKA=\sphericalangle DAK=\sphericalangle BAK$, so triangle $BAK$ is isosceles. Similarly, because $AB||DL$ we see that triangle $ADL$ is also isosceles. Hence: \begin{align*} DO^2-r^2=P(D, o)=DC\cdot DL=AB\cdot AD=BK\cdot BC=P(B, o)=BO^2-r^2, \end{align*} so $DO=BO$, thus $\Delta BOC\equiv \Delta DOL$. Therefore $\sphericalangle OBC=\sphericalangle ODL=\sphericalangle ODC$, so $OCDB$ is cyclic and we’re done.


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