# Concurrency involving two tangents and a secant of a circle

I found out this problem while fooling around with Geogebra.

Let $$(O)$$ be a circle in the plane, $$A$$ be a point lies outside of the circle. From $$A$$, draw two tangents $$AB, AC$$ to $$(O)$$ ($$B, C\in (O)$$). A line $$d$$ passing through $$A$$ ($$O\not\in d$$) intersects $$(O)$$ at two points $$E, F$$. $$BC$$ and $$EF$$ intersects at $$K$$; $$CE, CF$$ intersects the line $$AO$$ at $$X, Y$$ respectively.

a. Prove that $$CK, FX, EY$$ are concurrent.

b. Let $$D$$ be the intersection of $$BO$$ and $$(O)$$. $$DE, DF$$ intersect $$AO$$ at $$M, N$$ respectively. Prove that $$DK, EN, FM$$ are concurrent.

What I have proved so far:

• $$AEMC, AFCN$$; $$EMYF, EXNF$$ are inscribed.
• $$OM = ON$$.

Indeed, since $$\widehat{CEM} =\widehat{EBD}$$ and $$\widehat{OBE} = \widehat{BAM}=\widehat{MAC}$$ then $$AEMC$$ is inscribed. Similarly, $$AFCN$$ is inscribed.

Now since $$\widehat{AME} = \widehat{ECA}$$ and $$\widehat{ECA} = \widehat{EFC}$$ then $$\widehat{EMY}+\widehat{EFY} = \pi$$, thus $$EMYF$$ is inscribed. Similarly, $$EXNF$$ is inscribed.

Note that $$MNDC$$ is an isosceles trapezoid, then $$BMDN$$ is a parallelogram thus $$O$$ is the midpoint of $$MN$$.

Experimenting on Geogebra shows that the above results are true, however Still I can't prove them. Please help me, thanks.

UPDATE: I've edited my post to include my attempts. And thanks to Reza Rajaei and D S solutions, I still love to see a solution without using trigonometric Ceva theorem. Thanks for answering my questions.

• In addition, with the above configuration, can we prove that $IJ\parallel AO$? Mar 2 at 8:51
• that will take some work. Moreover, my Lemma 4 does not use the fact that $C$ and $D$ lie on the same circle, so it may be true, nor the fact that $D$ is the antpode of $B$.
– D S
Mar 2 at 8:53
• @anomino I have also proved $IJ || AO$ in an edit.
– D S
Mar 2 at 17:39

First, note that $$BECF$$ is a harmonic quadrilateral. This directly implies $$K = BC \cap EF$$ is the harmonic conjugate of $$A$$ w.r.t $$E,F$$.

Then note that if $$J = EY \cap FX$$, then $$CJ \cap AF$$ must be the harmonic conjugate of $$A$$ w.r.t $$E,F$$, by the construction of harmonic conjugate. Hence, $$CJ \cap AF = K$$.

The second part follows by a similar reasoning.

If the terminology is too technical, you can proceed using the following lemmas.

Lemma 1: Given tangents $$BA$$ and $$CA$$ to the circumcircle of $$\Delta BCE$$, point $$F \neq E$$ which is the intersection of $$AE$$ and the circumcircle, the following holds: $$\overline{BE}\cdot\overline{CF} = \overline{BF}\cdot\overline{CE}$$

This should be doable by sine rule alone.

Lemma 2: Given tangents $$BA$$ and $$CA$$ to the circumcircle of $$\Delta BCE$$, point $$F \neq E$$ which is the intersection of $$AE$$ and the circumcircle, the following holds: $$\frac{AE}{AF} = \left( \frac{AB}{AF}\right)^2 = \left(\frac{EB}{BF}\right)^2$$

This can be done by the power of point $$A$$ and sine rule.

Lemma 3: Given tangents $$BA$$ and $$CA$$ to the circumcircle of $$\Delta BCE$$, point $$F \neq E$$ which is the intersection of $$AE$$ and the circumcircle, and $$K = EF \cap BC$$, the following holds: $$\frac{KE}{KF} = -\left(\frac{EB}{BF}\right)^2$$

You will need the sine rule and lemma 1 here.

Lemma 4: Given three collinear points $$A, E, F$$ (not necessarily in that order, but here the order is the same), an arbitrary point $$T$$ outside this line, another line $$l$$ passing through $$A$$ such that $$l \cap ET = P$$ and $$l \cap FT = Q$$, let $$H = EQ \cap FP$$. Then, if $$TH \cap EF = K'$$, the following holds: $$\frac{AE}{AF} = -\frac{K'E}{K'F}$$

You will need Ceva's theorem and Menelaus's Theorem here.

Note that lemma 2 and 3 together imply $$\frac{AE}{AF} = -\frac{KE}{KF}$$ Then, for part 1, put $$l = AO$$, $$T = C$$, $$P = X$$, $$Q = Y$$, $$H = J$$, and in part 2, put $$l = AQ$$, $$T = D$$, $$P = M$$, $$Q = N$$, $$H = I$$, to conclude $$K = K'$$ in both cases. Done!

To prove $$IJ \mid \mid AO$$:

$$\measuredangle JEI = \measuredangle YEN = \measuredangle YEF - \measuredangle NEF = \measuredangle YMF - \measuredangle NXF = \measuredangle NMF - \measuredangle NXF = \measuredangle XFM = \measuredangle JFI$$ So that $$E,J,I,F$$ are concyclic.
Then $$\measuredangle EIJ = \measuredangle EFJ = \measuredangle EFX = \measuredangle ENX \implies IJ \mid \mid AO \tag*{\blacksquare}$$

• Your solution requires some technical knowledge. I have to review those Wiki pages! Mar 2 at 7:21
• All of them have simple proofs though, @RezaRajaei. Maybe I will add them.
– D S
Mar 2 at 7:46
• Lemma 4 uses only normal version of Ceva's theorem
– D S
Mar 2 at 8:51
• I checked out the wiki page [+1]. Just by using the definition and showing that $K$ is the harmonic conjugate of $A$, everything is clear. That was helpful. Of course, in spirit, both solutions are using sort of well-known concurrency theorems! Mar 2 at 8:52
• As an aside, we can show $IJ \parallel CD \parallel AO$ by Menelaus on the lines $EJY, EIN$ in the triangles $KCF, KDF$. We get that $KJ/JC = KE/EF \times FY/YC = KE/EF \times FN/ND = KI/ID$. Mar 2 at 23:13

For Part $$(a)$$:

Step $$1$$:

Show that: $$\frac{FY}{YC}=\frac{AF}{AC} \times \frac{\sin \angle YAF}{ \sin \angle YAC}.$$

Step $$2$$:

Show that: $$\frac{XC}{XE}=\frac{AC}{AE} \times \frac{\sin \angle YAC}{ \sin \angle YAF}.$$

Step $$3$$:

Show that: $$\frac{EK}{KF}=\frac{EB}{BF} \times \frac{\sin \angle CBE}{ \sin \angle CBF}=\frac{EB}{BF} \times \frac{CE}{CF}.$$

Step $$4$$:

Show that $$\frac{EB}{BF}=\frac{AB}{AF}$$ and $$\frac{CE}{CF}=\frac{AE}{AC}$$.

Step $$5$$:

Conclude that $$\frac{FY}{YC} \times \frac{XC}{XE} \times \frac{EK}{KF}=1$$. Now, note that the claim follows from Ceva's theorem.

For Part $$(b)$$:

Do exactly the same process.

• @DS You're right. However, I posted an incomplete solution. At least I tried to post hints and a general idea rather than a detailed solution. Maybe without Ceva's theorem, this could be a really hard problem. Mar 2 at 7:10
• this is a two-liner, I think I will post my solution now
– D S
Mar 2 at 7:12
• Thank you @RezaRajaei, I really love to see a solution without using trigonometric Ceva theorem. Mar 2 at 8:00
• @anonimo You're welcome. Mar 2 at 8:57
• @anonimo what is the reason for your aversion to the trigonometric form? You have included the tag contest math; there are many problems from various olympiads that either use it or its proof in disguise for all their solutions.
– D S
Mar 2 at 9:36