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By the Ceva theorem, applied to the triangle BCF and the concurrent cevians BM, CA and FE (in fact, these cevians concur at the point D), we have $\dfrac{MF}{MC}\cdot\dfrac{EC}{EB}\cdot\dfrac{AB}{AF}=1$ (we work with non-directed segments here). Hence, $\dfrac{MF}{MC}=\dfrac{AF}{AB}\cdot\dfrac{EB}{EC}=\dfrac{AF}{AB}: \dfrac{EC}{EB}$. Thus, MF = MC holds if and only if $\dfrac{AF}{AB}=\dfrac{EC}{EB}$. But by Thales, $\dfrac{AF}{AB}=\dfrac{EC}{EB}$ is equivalent to AE || FC, and obviously we have AE || FC if and only if < EAC = < ACF. Now, since the points A, B, D and E lie on one circle, < EAD = < EBD, what rewrites as < EAC = < CBM; on the other hand, trivially < ACF = < DCM. Thus, we have < EAC = < ACF if and only if < CBM = < DCM. Now, as it is clear that < CMB = < DMC, we have < CBM = < DCM if and only if the triangles CMB and DMC are similar (in fact, you can see it this way: the triangles CMB and DMC already have one pair of equal angles, namely < CMB = < DMC, and thus they are similar if and only if they have one more pair of equal angles). On the other hand, the triangles CMB and DMC are similar if and only if MB : MC = MC : MD (in fact, the triangles CMB and DMC have one pair of equal angles, namely < CMB = < DMC, and thus they are similar if and only if the sides adjacent to these angles are in the same ratio). Finally, it is clear that MB : MC = MC : MD is equivalent to $MB\cdot MD=MC^2$. Combining all these equivalences, we see that MF = MC holds if and only if $MB\cdot MD=MC^2$.

It's such a easy problem.Why you use such a difficult method to solve it.

MithsApprentice wrote: Let $ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E$, respectively. Lines $AB$ and $DE$ intersect at $F$, while lines $BD$ and $CF$ intersect at $M$. Prove that $MF = MC$ if and only if $MB\cdot MD = MC^2$. Assume $MF = MC$. 1) Draw parallel to $AD$ from $F$ and parallel to $ED$ from $C$. Let them intersect at $Y$. 2) $Y$ must be on line $BM$ because $FDCY$ is a parallelogram and its diagonals must intersect at their midpoints. 3) $\angle FYC = \angle ADE$ and $\frac{AD}{FY} = \frac{BD}{BY} = \frac{DE}{CY}$ therefore triangles $FYC$ and $ADE$ are similar and $FC \parallel AE$ 4) $\angle MCA = \angle YFC = \angle DAE = \angle DBE$ hence the circumcircle of triangle $BDC$ is tangent to line $MC$ therefore $MC^2 = MD \cdot DB$ Assume $MC^2 = MD \cdot DB$ 1) Then $\angle MCD = \angle DBE = \angle DAE$ hence $FC \parallel AE$ 2) $\angle MFE = \angle DEA = \angle DBA$ hence the circumcircle of triangle $BDF$ tangent to line $MF$ 3) Hence $MF^2 = MD \cdot MB = MC^2 \Rightarrow MF = MC$ Daniel

nice solution dhernandez wrote: 3) $\angle FYC = \angle ADE$ and $\frac{AD}{FY} = \frac{BD}{BY} = \frac{DE}{CY}$ therefore triangles $FYC$ and $ADE$ are similar and $FC \parallel AE$ 4) $\angle MCA = \angle YFC = \angle DAE = \angle DBE$ hence the circumcircle of triangle $BDC$ is tangent to line $MC$ therefore $MC^2 = MD \cdot DB$ You can also say that if FDCY is a parallelogramme, then FBCY is cyclic ( beause $\angle ADE = \angle FDC = \angle FYC$), so power of M: $MC^2= MF \cdot MC = MY \cdot MB = MD \cdot MB$

MithsApprentice wrote: Let $ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E,$ respectively. Lines $AB$ and $DE$ intersect at $F,$ while lines $BD$ and $CF$ intersect at $M.$ Prove that $MF = MC$ if and only if $MB\cdot MD = MC^{2}$ Nice problem! And here is a nice proof. Proof. Take $G\in BD: \,FG\parallel CD$ We have: $MF=MC\Longleftrightarrow \textrm{the quadrilateral}\; CDFG\; \textrm{is a parallelogram}\\ \Longleftrightarrow FD\parallel CG\Longleftrightarrow\angle FDA=\angle GCD\Longleftrightarrow\angle FDA+\angle CGF=180^\circ\\ \Longleftrightarrow \angle ABE+\angle CGF=180^\circ\Longleftrightarrow\textrm{the quadrilateral}\;CBGF\;\textrm{is cyclic}\\ \Longleftrightarrow\angle CBM=\angle CBG=\angle CFG=\angle DCF=\angle DCM\\ \Longleftrightarrow\triangle BCM\sim\triangle CDM\Longleftrightarrow MB\cdot MD=MC^{2}$

If MC=MF. Assume that AE not paralel to CF, call T is intersection of AE and CD. we have (CFMT) = -1, on the other hand. M is midpoint of CD. so nonsense. Hence, CF // AE, easy to finish problem If MC.MC = MB. MD, easy

If $ MC=MF$. Assume that $ AE$ not paralel to $ CF$, call $ T$ is intersection of $ AE$ and $ CD$. we have $ (CFMT) = -1$, on the other hand. M is midpoint of CD. so nonsense. Hence, $ CF // AE$, easy to finish problem If $ MC^{2} = MB. MD$, easy

Sorry for bringing up an old topic. How did the solvers know to create the parallelogram. Is this just a well known construction? Does my solution work? Solution Assume that $ MC=MF$. Now, by Ceva's theorem $ \frac{MC}{MF} \cdot \frac{BE}{CE} \cdot \frac{FA}{BA}=\frac{BE}{CE} \cdot \frac{FA}{BA}=1$. Therefore $ \frac{BE}{CE} = \frac{BA}{FA}$ and by SAS similarity, $ BAE \sim BFC$. Therefore $ AE \| FC$. Now note that $ \angle{MFD}=\angle{DEA}=\angle{DBA}$. Hence by AA similarity, $ MDF \sim MFB$ and it follows that $ \frac{MD}{MF}=\frac{MF}{MB}$. Rearranging yields that $ MF^2=MC^2=MD \cdot MB$. Assume that $ MC^2=MD \cdot MB$. It follows that $ \frac{MD}{MC}=\frac{MC}{MB}$ which by SAS implies that $ MCD \sim MBC$. Hence $ \angle{MCB}=\angle{MDC}=\angle{ADB}=\angle{AEB}$ and therefore $ FC \| AE$. This implies that $ AEB \sim FCB$ and that $ \frac{BE}{CE} = \frac{BA}{FA}$. By Ceva's theorem, it follows that $ \frac{MC}{MF} \cdot \frac{BE}{CE} \cdot \frac{FA}{BA}=\frac{MC}{MF}=1$. Therefore $ MC=MF$.

Quite a nice problem. It is possible to prove both parts simultaneously. By Ceva's Theorem, $\frac{FM}{MC} \cdot \frac{CE}{EB} \cdot \frac{BA}{AF} = 1$ so $MF=MC \Leftrightarrow \frac{CE}{EB} \cdot \frac{BA}{AF} = 1$ $\Leftrightarrow \frac{CE}{EB} = \frac{FA}{AB}$ $\Leftrightarrow EA || CF$ $\Leftrightarrow \angle MCA = \angle DAE$ $\Leftrightarrow \angle MCA = \angle DBE$ $\Leftrightarrow MC^2 = MD \cdot MB$ where the last step follows from the Tangent-Secant Theorem and its converse.

First let us prove the if statement. If $MD\cdot MB=MC^2$, then $MC$ is tangent to $\odot DBC$ at $C$ and thus $\angle MCD=\angle MBC$ and since quadrilateral $ABDE$ is cyclic, we have $\angle MBC=\angle FAD$ so $AF||FC$ so $\triangle BAE~\triangle BFC$ thus $\frac{BA}{AF}=\frac{BE}{EC}\implies \frac{BA}{AF}\cdot\frac{EC}{BE}=1$. Now by Ceva's, we have \[\frac{FA}{AB}\cdot\frac{BE}{EC}\cdot\frac{CM}{MF}=1\] and using $\frac{BA}{AF}\cdot\frac{EC}{BE}=1$, we have $\frac{CM}{MF}=1\implies CM=MF$ as desired. Now to prove the only if statement, we begin with $MF=MC$ now by Ceva's we find $\frac{FA}{AB}\cdot\frac{BF}{FC}=1\implies{FA}{AB}=\frac{FC}{BF}$ SO $\triangle BAE~\triangle BFC$ which means $AF||FC$ and so $\angle DBF=\angle FAD=\angle DCM$ so $MC$ is tangent to $\odot BDC$ at point $C$ which means $MC^2=MD\cdot MB$ by Power of a Point.

We work in the projective plane. Let $X = AE \cap FC$ First, let us show that if $MF = MC$, then $MC^2 = MD\cdot MB$. Note that $(F,C;M,X)$ is a harmonic division. Because $MF = MC$, we must have that $X$ is the point at infinity on line $FC$, so we have that $FC$ is parallel to $AE$. Thus, we have that $\angle{FCA} = \angle{CAE} = \angle{DBC}$, where the last angle equality follows from cyclic quadrilateral $ADEB$. Thus, we see that $MC$ is tangent to the circumcircle of $\triangle{BDC}$ at $C$, and the result follows. Now let us show that if $MC^2 = MD\cdot MB$, then $MF = MC$. Note that we must have that $\angle{MCA} = \angle{EBD} = \angle{DAE}$, so we get that $AE$ is parallel to $FC$ again. Thus, $X$ is the point at infinity on line $FC$, but we still have that $(F,C;M,X)$ is harmonic, so we must have that $M$ is the midpoint of $FC$, so we are done. $\blacksquare$ This solution is essentially the same as the ones above, however, a lot of the work using ceva is encapsulated in a well know harmonic division lemma.

First I do ''only if'' part ---- Note that , in triangle $BCF$ , , we have $BM,EF,AC$ as 3 concurrent cevians. hence , by using ceva's theorem , we get that $AE$ is parallel to $CF$. hence , after a little bit of anngle chasing , we get $\Delta BCM$ is similar to $\Delta CDM$. hence , $MB.MD=MC^2$. for if part -- $MB.MD=MC^2$ implies that , $\Delta BCM$ is similar to $\Delta CDM$ with $\angle CBM=\angle DCM$ , and $\angle CDM= \angle BCM$ hence , a little angle chasing shows that $AE$ is parallel to $BC$. hence , by Ceva's theorem , we get that $MC=MF$

Suppose $M$ is the midpoint,so from ceva we have $AE$ and $CF$ are parallel,so $\angle {DAE}=\angle {DCF}$,but $\angle {DAE}=\angle {DBE}$,hence $\angle {DBE}=\angle {DCM}$,so $MC$ is tangent to the circle $DBC$,so $MC^2=MB.MD$,hence done.

i think the question was too easy for usamo.. ceva's theorem with a little angle chasing gives quite an easy proof...

Can someone explain how to actually draw the diagram? 5 sheets of paper later my diagrams are still wacky..

Notice that by Power of a Point \begin{align*} MB \cdot MD = MC^2 &\iff MC \; \text{is tangent to} \odot (BCD) \\ &\iff \measuredangle MCB = -\measuredangle MDC \\ &\iff \measuredangle FCE = -\measuredangle MDC \\ & \qquad \qquad \qquad \! = -\measuredangle BDA = -\measuredangle BEA = -\measuredangle CEA = \measuredangle AEC \end{align*} where the angles are directed, and we have used the well-known "Tangent-Angle Theorem." However, $\measuredangle FCE = \measuredangle AEC$ is equivalent to $CF \parallel AE.$ Now, let $O$ be the center of $\odot (ABED)$ and denote $X = AE \cap BD.$ Then by Brokard's Theorem on cyclic quadrilateral $ABED$, we deduce that $O$ is the orthocenter of $\triangle CFX.$ Therefore, $OX \perp CF.$ It follows that \[MB \cdot MD = MC^2 \iff CF \parallel AE \iff OX \perp AE.\] Notice that since $\overline{AE}$ is a chord of a circle centered at $O$, and $X \in \overline{AE}$, we have $OX \perp AE \iff X$ is the midpoint of $\overline{AE}.$ Then since \begin{align*} MB \cdot MD = MC^2 &\iff CF \parallel AE \quad \text{and} \\ MB \cdot MD = MC^2 &\iff X \quad \text{is the midpoint of} \quad \overline{AE} \end{align*} we deduce that \[MB \cdot MD = MC^2 \iff M = BX \cap CF \quad \text{is the midpoint of} \quad \overline{CF}\] as desired. $\square$

hamup1 wrote: Can someone explain how to actually draw the diagram? 5 sheets of paper later my diagrams are still wacky.. Drawing a diagram on this one is pretty tough . I think you just have draw accurately and hope for a decent figure. One bit of advice though is if you construct point $M$ and $M$ turns out to lie outside of $\overline{CF}$ (which would be bad, since $M$ is supposed to be the midpoint of $\overline{CF}$), you can switch points $A$ and $B$ in your diagram, and then $M$ should end up on $\overline{CF}.$

I had such an unorthodox solution. I botched up the only-if part so I didn't include it. Is my solution right?

I can't follow your angle chasing just from the diagram lol but I think you should be fine. It is easier to show that $MF$ is tangent to $(BDF)$ .

By Ceva's theorem on $\triangle BCF$ and cevians $BM,CA,FE$, $CM=MF\iff AE\parallel CF.$ Also, $AE\parallel CF\iff \angle MCD=\angle DAE.$ Since $\angle DAE=\angle DBE$, $\angle MCD=\angle DAE\iff CM$ tangent to $(BCD)\iff MC^2=MD\cdot MB$ as desired.

Assume $MF=MC$ connect $AE$ by ceva, $FA/AB * BE/EC * MC/MF = 1$ $FA/AB = EC/BE$ $AE//FC$ $AED$ similar to $CDF$ Because $BADE$ is cyclic, $\angle DBE=\angle EAD=\angle FCD$ So $BMC$ is similar to $DMC$ and $MB*MD=MC^2$ Assume $MB\cdot MD = MC^2$ Get that $BMC$ is similar to $DMC$ So $\angle DBE=\angle FCD$ and $\angle DBE = \angle EAD$ (cyclic quadrilateral) $AE//FC$ $AF/AB=EC/EB$ By ceva, $FA/AB * BE/EC * MC/MF = 1$ $EC/EB * BE/EC * MC/MF = 1$ $MC/MF = 1$ $MC=MF$ first oly geo solve

IAmTheHazard wrote: First we prove the following lemma: Lemma: In $\triangle ABC$ with concurrent cevians $AD,BE,CF$, we have $EF\parallel BC$ if and only if $BD=CD$. Proof: Using Ceva's and similar triangles we have: $$BD=CD \iff \frac{BF}{AF}\cdot \frac{AE}{CE}=1 \iff \frac{BF}{AF}=\frac{CE}{AE} \iff \frac{AB}{AF}=\frac{AC}{AE} \iff \triangle AEF \sim \triangle ACB \iff EF \parallel BC$$as desired. We note that the condition $MB\cdot MD=MC^2$ can be rewritten as $\frac{MB}{MC}=\frac{MC}{MD}$, after which is becomes clear that we have: $$MB\cdot MD=MC^2 \iff \triangle MBC \sim \triangle MCD,$$so it is equivalent to prove: $$\triangle MBC\sim \triangle MCD \iff MF=MC.$$Now observe that we have (angles are directed): $$\measuredangle FCB=\measuredangle MCE~ \text{ and } ~\measuredangle CDM=\measuredangle ADB=\measuredangle AEB.$$Noting that $\triangle MBC\sim \triangle MCD \iff \measuredangle MCE=\measuredangle CDM$ and $\measuredangle MCE=\measuredangle CDM \iff EA \parallel CF$, we see that it is equivalent to show that: $$MF=MC \iff EA \parallel CF,$$which is a direct application of our lemma. $\blacksquare$ I did this problem again because why not Note that $$MB\cdot MD=MC^2 \iff \overline{MC} \text{ tangent to } (BDC) \iff \measuredangle MCD=\measuredangle CBD \iff \measuredangle FCA=\measuredangle EAC \iff \overline{AE} \parallel \overline{CF} \iff MF=MC,$$where the last step follows by Ceva. $\blacksquare$

Note that $MF = MC$ is equivalent by Ceva's theorem to $\frac{BA}{AF} = \frac{CE}{BC}$, or that $AE \parallel FC$ by similarity. This is in turn equivalent to \[ \measuredangle DBC = \measuredangle DBE = \measuredangle DAE = \measuredangle DCF \]which is just the same as $(BCD)$ being tangent to line $FC$, which is the power of the point condition.

Beautiful problem! We have MD/MC=MC/MB, and also <BMC=<CMD. Then triangles CMD and BMC are similar. Also, note that <AEB=<ADB=<CDM, so by AA (angle CBM is an extra) triangles BEQ and BCM are similar, where Q is the intersection of AE and BM. This is equivalent to EA parallel to BC, whence BA/BF=BE/BC. Now, we note that if a/b=c/d then a/(b-a)=c/(d-c) since (b-a)/a=(d-c)/c is equivalent to b/a-1=d/c-1. Applying this we get BA/(BF-BA)=BE/(BC-BE) and hence Ceva's says that BE/BC*CM/MF*FA/BA=1, which upon simplifying yields the desired result. Each of these steps are reversible (namely we would obtain SAS similarity on BAE and BFC and then MDC~QEB~MCB by AA), so we're done! $\blacksquare$ Alternatively we can continue as follows: $MB\cdot MD=MC^2 \iff \overline{MC} \text{ tangent to } (BDC) \iff \measuredangle MCD=\measuredangle CBD \iff \measuredangle FCA=\measuredangle EAC \iff \overline{AE} \parallel \overline{CF} \iff MF=MC,$ where the last follows from Ceva's as I showed before. That angle chasing of <MCD=<CBD could then be done this way, or we could FIRST angle chase and see this. Anyways, I saw this after looking at it for a bit longer. This is a much simpler solution. Remarks. Strange, at the beginning I thought it would either have something to do with constructing a tangent MG to show MG=MC isosceles triangle, and conjectured it would lie on the perp. bisector of AB, but apparently not, although close.. someone confirm if my diagram is then wrong (yes geogebra COULD be wrong bc note I have two points H and M, H is the actual midpoint so I tried to make the config as close as possible, since theres no way to ACTUALLY do that) or is this just not true? I also had an idea of using radical axis, seems like suggesting M on radical axis of circle with circle of radius 0. I'd like to see solutions with that (if even possible), but I'll first post mine so that I don't get spoilered.

By Ceva's, $MC=MF$ iff $\tfrac{BA}{BF}=\tfrac{BE}{BC}$, i.e., $AE\parallel CF$, which in turn is equivalent to $\angle DCM=\angle DAE=\angle DBE$. But $\angle DCM=\angle DBE$ iff $(BCD)$ is tangent to $CF$ at $C$, which by PoP is equivalent to $MB\cdot MD=MC^2$ as desired.

Note that \[MF=MC\Longleftrightarrow\frac{BA}{AF}=\frac{BE}{EC}\Longleftrightarrow AE\parallel FC\Longleftrightarrow\angle FCB=\angle AEB=\angle ADB=\angle CDM\]\[\Longleftrightarrow MC\text{ is tangent to }(DCB)\text{ at }C\Longleftrightarrow MC^2=MD\cdot MB\]

MithsApprentice wrote: Let $ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E$, respectively. Lines $AB$ and $DE$ intersect at $F$, while lines $BD$ and $CF$ intersect at $M$. Prove that $MF = MC$ if and only if $MB\cdot MD = MC^2$.

$\color{magenta} \boxed{\textbf{SOLUTION P4}}$ $\color{red} \textbf{Geo Marabot Solve 7}$ By $\textbf{Ceva's Theorem,}$ $$MF=MC \Longleftrightarrow \frac{FA}{AB}=\frac{CE}{EB} \Longleftrightarrow CF\parallel AE \Longleftrightarrow \angle FCA=\angle CAE=\angle DBC \Longleftrightarrow MB \cdot MD=MC^2\blacksquare$$

Note that Ceva's theorem yields $\overline{AE} \parallel \overline{CF}$. Hence, \[\angle MBC = \angle DBE = \angle DAE = \angle MCD,\] which implies $\triangle MBC \sim \triangle MCD \iff MB^2 = MC \cdot MD$. All the other steps are reversible, so we are done. $\square$