# Are points H, M, K' collinear?

In solving this problem, I am having some trouble in convincing others on the final step. Here is the description of that problem:-

Let △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 $DT \bot MD$ and $ET \bot ME$.

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

With all the given, the object is to prove OB = OC. Through the construction of point X and Y on AB and AC respectively such that AB = BX and AC = CY, we know that XZ = ZY, by midpoint theorem.

The job is done if I can show TX = TY. Or equivalently, if I can show that $\angle TZY = 90^0$. This means I have to show AHZK’ is a rectangle. In turn, I have to show that H, M, K’ are collinear.

I tried to use Pascal’s theorem and radical axis theorem. The argument seems to be not that convincing. Can some verify my approaches? Or give comment to make my argument more convincing? Of course, more simpler or elegant (geometric) methods are also welcome.

• I think I have nailed it. Give me an hour to write the proof in detail. – Oldboy Jul 3 '18 at 13:28

Let $X$ be such that $M$ is the midpoint of $AX$. Then $AX$ is a diameter of the circumcircle of $ADE$. It follows that $\angle CEX = \frac \pi 2$. We also have $AB \parallel CX$ so $\angle XCE = \angle BAC$.

Now, $\angle MET = \frac \pi 2$ and $$\angle TME = \frac 12 \angle DME = \frac 12 \cdot 2\angle DAE = \angle BAC = \angle XCE.$$ It follows that $\triangle CEX \sim \triangle MET$ because these triangles have equal angles. Therefore $$\frac{CE}{EX} = \frac{ME}{ET}$$ which along with $\angle CEM = \frac \pi 2 - \angle MEX = \angle XET$ gives $\triangle XET \sim \triangle CEM$ by SAS. Thus $\angle MCE = \angle TXE$. This means that $X,E,C,Y$ are concyclic, where $Y$ is the common point of $TX$ and $BC$. Since $\angle CEX = \frac \pi 2$, we have $\angle XYC = \frac \pi 2$. Thus $TX \perp BC$. Since $M$ is the midpoint of $AX$ and $O$ is the midpoint of $AT$, we have $OM \parallel XT$. Therefore $OM \perp BC$ and since $M$ is the midpoint of $BC$, we have $OB=OC$ as desired.

Many thanks to Oldboy for providing the picture. An interactive version is also available: http://geogebra.org/classic/bzu4sts4

• One small correction: $\triangle XET \sim \triangle CEM$. There is a similar solution here. – Maxim Jul 7 '18 at 13:58
• Nice solution. Can you explain why $TX \perp BC$ in geometric terms (rather than using same orientation)? – Mick Jul 7 '18 at 13:58
• @Maxim, thanks, I'll correct this in a moment. – timon92 Jul 7 '18 at 14:12
• @Mick This follows from an easy angle chasing. – timon92 Jul 7 '18 at 14:14
• Excellent, this one should be accepted. I have also added a nicely colored GeoGebra drawing here, feel free to add it to your answer: geogebra.org/classic/bzu4sts4 – Oldboy Jul 8 '18 at 10:57

Lemma #1 Let HRLI and ARLJ be two intersecting circles with HRA and JLI being straight lines. Then, AJ // HI. Proof is skipped

Lemma #2 Let A, R, L be inscribed in a circle with AL as the diameter and O as the center (i.e. $\angle ARL = 90^0$). $\triangle RLJ$ is similarly defined with $\angle RLJ = 90^0$. Then, ARLJ is cyclic. The proof is skipped.

As mentioned earlier, since XZ = 2BM = 2MC = ZY, we only need to show $TZY =90^0$.

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

We have three intersecting circles and they are (1) The red circle M (diameter = AMZ); (2) The blue circle B (diameter = ABX); and (3) The cyan circle C (diameter = ACY). They share the same common chord AH. With respect to AH, CMB being the line of centers, will bisect AH perpendicularly at R. That is, CRMB is a straight line.

By angles in semi-circle, $\angle YHA = \angle AHZ = \angle AHX = 90^0$. This in turn means XZHY is a straight line.

The orange circle O (diameter = AOT) cuts circle M at A and K. That is, AK is their common chord with GMOU being the corresponding line of centers (where G and U are points as shown). By angle in semi-circle, $\angle AKZ = \angle AKT = 90^0$. This, in turn, means TZK is a straight line.

(Please skip the next two paragraphs.)

Draw the green circle (centered at B, radius = BZ). Produce ZB to cut the same circle at J. That is, Z(B)J is the diameter. Note that $\angle ZKJ = 90^0$ and AKJ is a straight line. Then, BK = BZ. The above, together with MK = MZ, implies MKBZ is a kite with $\angle MNK = 90^0$.

Finally, KNMG, GARM, MRHU, and UMNZ are rectangles in turn. The required result follows.

Produce ZB to cut AK produced at J. Draw JI perpendicular to XZ cutting X(ZUH)Y at I. Clearly, ZKJI is cyclic. Produce C(RMN)B to cut IJ at L. Since L is the 4th vertex of the potential rectangle IHRL, $\angle BLJ = \angle BLI = 90^0$. In addition, IHRL is cyclic.

Form the circle passing through A, R, L with diameter = AL. Also, form the circle passing through R, L, J with diameter = RJ. (Some conditions need to be added before Lemma 2 can be applied, ARLJ is cyclic. TO BE FIXED). By lemma 1, AJ // HI.

The required result follows.

• @Maxim This is another shot at the problem. Please check. – Mick Jul 5 '18 at 16:23
• Why is $\angle ZKJ = \pi/2$? We do not know yet that the green circle passes through $K$. [Apparently SE allows you to notify only the users who have commented on the current answer. Your comment wasn't sent to my inbox.] – Maxim Jul 5 '18 at 17:53
• @Maxim Sorry for making a confusion. I thought that is the right place to send a message to you so that you don't need to search. Try to fix the error. Introduced two lemmas to simplify my explanation a bit. – Mick Jul 7 '18 at 11:41
• Perhaps something else is intended here, but Lemma 2 is false as stated. You can extend the rays $RA$ and $LJ$ indefinitely. – Maxim Jul 7 '18 at 13:04
• @Maxim I have the wording of lemma 2 changed. Then, I need to add the condition of the crossing center. To be fixed. – Mick Jul 7 '18 at 13:31