Show that $OB=OC$ [closed]

Let $\triangle ABC$ and $M$ be the middle of $[BC]$.

Let $D\in AB$ with $B \in [AD]$ and $E \in AC$ with $C \in [AE]$ such that $AM=MD=ME$.

Let $T$ such that $DT \perp MD$ and $ET\perp ME$.

If $O$ is the middle of $AT$ show that $OB=OC$.

This is my picture:

closed as off-topic by user21820, Gibbs, Claude Leibovici, Namaste, JonMark PerryJun 7 '18 at 11:41

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• I have just uploaded a revised version which is, independent of the controversial issue that we have discussed, and far simpler than I have thought. – Mick Jun 3 '18 at 13:00
• Your post was closed. Before that, I have posted my initial solution in math.stackexchange.com/questions/2798092/… for help. Now, I can only show my final version in there. Do comment generously. – Mick Jun 8 '18 at 1:24

Just bash it with the coordinate method. Wlog, let $A = (x_A, y_A), B = (-1, 0), C = (1, 0)$. The coordinates of $D$ and $E$ are rational fractions, because one intersection point of the chords $AB$ and $AC$ with the circle is already known. The coordinates of $T$ come out as $$T = \left( -x_A, \frac {y_A (1 + x_A^2 + y_A^2)} {1 - x_A^2 - y_A^2} \right).$$

We only need $x_T = -x_A$ to show that $O$ is on the $y$ axis.

There are no constraints except $x_A^2 + y_A^2 \neq 1$; $D$ can be on the ray $AB$, or on the segment $AB$, or on the ray $BA$.

deleted Jun 10 '18 at 4:12
• Thank you! I am looking for a synthetic proof because it is from a contest for children. – rafa May 23 '18 at 12:15

Denote by $\mathscr C$ the circle $(M, MA)$. Let $F$ be the point diametrically opposite to $A$ on $\mathscr C$. Then $\angle FDB = 90^{\circ}$. Let $\angle DBF=\alpha$. $BACF$ is a parallelogram because $BM=MC$ and $AM=MF$. Thus, $\angle DAE=\alpha$. Clearly, $\triangle DMT=\triangle EMT$. So $\angle DMT=\frac 1 2 \angle DME=\angle DAE=\alpha$. We have shown that $\triangle DBF \sim \triangle DMT$. Now consider the rotational homothety with the center at $D$ that sends $T$ to $M$ (its angle is $90^{\circ}$ and its factor is $\cot \alpha$). Clearly, it also sends $F$ to $B$. So it sends the segment $FT$ to the segment $BM$ and $FT \perp BM$. Since $MO$ is a midline of $\triangle FAT$, $MO \parallel FT$ and $MO \perp BM.$ Thus, $\triangle BMO=\triangle CMO$.

deleted Jun 10 '18 at 4:12

The logic is quite simple because, by midpoint theorem, we only need to show TX = TY (see below). However, my solution slightly complicated. I hope someone can provide a more elegant version.

Clearly, M is the center of the red circle ADE. Produce AM to cut that circle at Z such that AMZ is a diameter of that circle. Therefore, $\angle AEZ = 90^0$.

Through Z, draw XZY // BMC cutting ABD produced and ACE produced at X and Y respectively. By intercept and midpoint theorem, $XZ = 2 BM = 2 MC = ZY$ … (1).

Note that AB = BX, and AC = CY. Note also that CBZY is a //gm with $\angle CBZ = \angle CYZ$.

Produce TZ to cut BMC at U. After adding the perpendiculars CH and HK, it is not difficult to see that all the red marked angles are equal, and $\angle B = \angle Y = \omega = \omega’$. Therefore, $\omega’ + \theta’ = 90^0$. … (2)

(1) and (2) imply UZT is the perpendicular bisector of XZY. Then, TX = TY.

Finally, $OB = \dfrac 12 TX = \dfrac 12 TY = OC$.

deleted by owner May 25 '18 at 5:09
• Can you explain why that angles are congruent? I tried to prove but I didn't succeed – rafa May 19 '18 at 19:55
• @rafa In my earlier version, I did use the word "congruent". In this version, showing the triangles are similar will be sufficient to yield the required result. From $\triangle KYC \sim \triangle HBC \sim \triangle EZY$, we get $\theta = \theta' = \theta''$. – Mick May 20 '18 at 11:11
• Why $\omega=\omega'$? – rafa May 20 '18 at 12:31
• @rafa I have just uploaded a newer version. Hope that can clarify the matter. – Mick May 23 '18 at 3:56

This is the 2nd version.

Extend AB, AC, AM to X, Y, Z respectively such that AB = BX, AC = CY and AM = MZ. By midpoint theorem, (1) XZ = 2BM = 2MC = ZY; (2) XZY is a straight line; and (3) BCYZ is a //gm.

Draw $AK’ \bot XZY$. Let AK’ cut BC at V. Then, CVMB is the perpendicular bisector of AK’.

Produce TZ to some point H’ such that $\angle AH’Z = \angle AH’T = 90^0$.

Draw the red circle ADZE (centered at M, radius = MA = MD = ME = MZ, diameter = AMZ). Note that H’ and K’ are con-cyclic points of that circle too.

Together with AZ = … = H’K’ (See the added below), we can say that AH’ZK’ is a rectangle. This further means TZH’ is the perpendicular bisector of XY. Hence, TX = TY. Result follows.

The perpendicular bisector of the common chord AH' will (1) cut the red circle at I and J; and (2) will pass through MJ, the line of centers. The selected diagonals (AZ, IJ and H'K') from the hexagon IAK'JZH' will concur at M, according to the PASCAL's theorem. This solves the collinearity problem of H', M, K'.

Another method

Imagine that $\omega_1, \omega_2, \omega_3$ are three duplicates of the red circle.

Let AZ, IJ and H’K’ (at this instant, it probably will not pass through M) be respectively the common chords of those three circles taken two at a time. These common chords are actually the three radical axes of the system. By radical axis theorem, they will meet at one single point (which must be M). Hence H’, M, K are collinear.

deleted Jun 10 '18 at 4:12
• @Mike Why is AH'ZK'a rectangle? – rafa May 23 '18 at 4:33
• @rafa The diagonals are equal in length. – Mick May 23 '18 at 4:36
• Why? We don't know that $K', M, H'$ are collinear – rafa May 23 '18 at 4:38
• But it's not true that the main diagonals are concurrent in any cyclic hexagon. Pascal's theorem doesn't say that. – Maxim May 25 '18 at 10:29
• Your second method again implies that any three chords in a circle are concurrent. This is clearly not true. The radical axis is undefined for concentric circles. – Maxim May 26 '18 at 9:31

This is the 3rd version.

ABD, ACE, and AM are respectively extended to X, Y, and Z such that AB = BX, AC = CY, and AM = MZ.

By midpoint theorem, XZ = 2BM = 2MC = ZY and XZY is a straight line parallel to BMC.

Since, by midpoint theorem, $OB = \dfrac 12 TX$ and $OC = \dfrac 12 TX$, we need to prove that TX = TY. Adding the fact stated above, we only need to show $\angle TZY = ….= 90^0$.

The red circle (centered at M, radius = AM, diameter = AMZ) will cut XZY at G such that $\angle AGZ = 90^0$.

The green circle (centered at O, radius = AO, diameter = AOT) will cut TY at K such that $\angle AKT = 90^0$.

The purple circle (centered at C, radius = AC, diameter = ACY) will cut TY at K also such that $\angle AKY = 90^0$.

The blue circle (centered at B, radius = AB, diameter = ABX) will cut XZY at G such that $\angle AGZ = 90^0$.

Note that the grey circle can also be formed (because $\angle MET = \angle MDT = 90^0$) passing through M, E, T, D with MT as diameter. Then, when TZ is produced, it will cut MC at H such that $\angle MHT = 90^0$ (because $\angle MHT$ subtends the diameter MT).

Note that AG is the chord common to circles C, M, B. CHMB is then a straight line (the line of centers) and it will cut the common chord AG perpendicularly at G’.

Result follows because Z is the fourth vertex of the rectangle HG’GZ.

deleted Jun 10 '18 at 4:12
• I think it needs to be proved that $TZ \perp MC$. – Maxim Jun 3 '18 at 17:24
• I have replaced the old picture with a new one which has the grey circle clearly shown. That will clarify why $TZ \perp MC$. – Mick Jun 4 '18 at 3:46
• You have replaced $TZ \perp MC$ with an equivalent statement "$TZ$, $MC$ and the circle on the diameter $MT$ intersect at the same point". This latter statement still needs to be proved. – Maxim Jun 4 '18 at 11:16
• @Maxim The grey circle cuts CN at H. That is, H is a con-cyclic point of the circle MDTE (whose diameter is MT). This makes $\angle T(Z)HM = 90^0$. In other words, $MC \perp TZ$ (extended). – Mick Jun 4 '18 at 13:17
• How do you define the point $H$? If $H$ is defined as the intersection point of $TZ$ and $MC$ (as in your answer), you need to prove that $H$ lies on the circle with the diameter $MT$. If $H$ is defined as the intersection point of the circle with the diameter $MT$ and $MC$ (as in your comment; I assume you meant $CM$, not $CN$), then you need to prove that $H$ lies on $TZ$. You cannot just assume that those two definitions give the same point. – Maxim Jun 4 '18 at 13:43