# Olympiad level | Similar Triangles

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 $$D$$.

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?

• I also notice $\angle{BKA}=\angle{CKL}$ as they're verticle angles. – user683941 Jun 22 '19 at 2:07
• 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. – 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!

• Oops, I just edited the post and changed the diagram cause I was working on it. Didn't notice you posted an answer – redsunx Jun 22 '19 at 6:57
• @MC Urist It's OK. I added previous points in my proof. :) Thank you that you restored it. – Michael Rozenberg Jun 22 '19 at 7:03
• 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... – redsunx Jun 22 '19 at 7:12
• @MC Urist I think, it's a very nice problem. – Michael Rozenberg Jun 22 '19 at 7:14
• 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. – 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.

(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.