Concurrency involving two tangents and a secant of a circle

Question

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.

Answer 1

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.

Answer 2

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, $CJ \cap AF = K$.

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The second part follows by a similar reasoning.

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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!

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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 $$\measuredangle EIJ = \measuredangle EFJ = \measuredangle EFX = \measuredangle ENX \implies IJ \mid \mid AO \tag*{$\blacksquare$}$$