Baltic Way 2019 P11 wrote: Let $ABC$ be a triangle with $AB = AC$. Let $M$ be the midpoint of $BC$. Let the circles with diameters $AC$ and $BM$ intersect at points $M$ and $P$. Let $MP$ intersect $AB$ at $Q$. Let $R$ be a point on $AP$ such that $QR \parallel BP$. Prove that $CP$ bisects $\angle RCB$. Solution: Let $AP \cap BC=E$ and Let $D$ be orthocenter WRT $\Delta AEC$ $\implies$ $PEMD$ is cyclic. Also, $\angle EPM$ $=$ $\angle ACB$ $=$ $\angle QBE$ $\implies$ $QBEP$ & $AQPD$ are cyclic. Since, $\angle QPB$ $=$ $90^{\circ}$ $\implies$ $QE||AM$. Let $H$ be orthocenter WRT $\Delta BMQ$. Clearly, $HEMD$ is rectangle. Note: $\angle AQR$ $=$ $\angle ABP$ $=$ $\angle QER$ $\implies$ $AQ$ is tangent to $\odot (QER)$. Let tangent to $\odot (QER)$ at $E$ intersect $\odot (BEPQ)$ at $F$ $\implies$ $BFQE$ is rectangle & $FE||AC$. Hence, $F$ $\in$ $PC$. Therefore, $\angle EFP$ $=$ $\angle EQP$ $=$ $\angle PED$ or $\angle FED$ $=$ $90^{\circ}$. Since $\angle DQB$ $=$ $\angle FQE$ $=90^{\circ}$ $\implies$ $\angle HQD$ $=$ $\angle FQB$ $=$ $\angle FEB$ $=$ $\angle HED$ $\implies$ $QH=HE$ or $RP=PE$ $\qquad \blacksquare$

So we claim the same, and slightly extend the results: Let $ABC$ be a triangle with $AB = AC$. Let $M$ be the midpoint of $BC$. Let the circles with diameters $AC$ and $BM$ intersect at points $M$ and $P$. The circle $(BMP)$ intersects for the second time $AB$ in a point $S$. The circle $PEC$ intersects for the second time $AC$ in a point $L$. Let $MP$ intersect $AB$ at $Q$. Let $R$ be a point on $AP$ such that $QR \parallel BP$. Let $E$ be the intersection $AP\cap BC$. The orthocenter $D$ of $\Delta AEC$ is in notation the intersection $AM\cap EL\cap CP$. Prove the following statements: $QE\| AM$ and $RL\| PC$ . $CP$ bisects $\angle RCB$. Proofs: Some general facts: We use the letter $y$ for the (measure of the) angle $\widehat {PAM}$, and try to compute all needed angles from th epicture using it together with (the measure of) $\widehat{MAC}$, which will be denoted by $\frac A2$ in the sequel. Then we have the situation from the picutre. Some small comments for the marked equal angles. - The angles marked by $\displaystyle \frac A2$ are equal, the two "inside $APMC$" in $A,P$ because $APMC$ is cyclic, the two "inside $\Delta PBC$" because adding a right angle we obtain $\widehat {BPC}$, and the two in $P,R$ by correspondence w.r.t. the parallels $PQ\| PB$. - The angles marked with $y$ in $A,C$ are equal because $APMC$ is cyclic, because of the right angles in $M,P$. The two $y$-angles in $A,B$ are equal, because adding $\widehat {BAP}$ we obtain the same $\frac 12 A$ angle (interior in $\widehat {BAM}$, respectively exterior $\widehat{BPE}$ w.r.t. the triangle $\Delta ABP$). The two angles marked $y$ in $B,Q$ are equal because of the correspondence w.r.t. $PQ\|BP$. - The blue angles are equal (in measure), because (in measure) $$ \begin{aligned} \widehat{AEC} &= 180^\circ - \widehat{EAC} - \widehat{ACE} = 180^\circ - (y+\frac {\hat A}2)-\hat C \\ &= 180^\circ-90^\circ - y = \\ &= 180^\circ-(90^\circ + y) = 180^\circ-\widehat{PQA} \\ &= \widehat{BQM} \ . \end{aligned} $$- So the quadrilateral $BEPQ$ is cyclic. So we can finally also identify the angle in $E$ in this quadrilateral as being $\widehat {PEQ}=\widehat {PBQ}=y$. This gives $$QE\|AM $$alfter comparing the green interior angles built with the secant line $AE$. - In particulat $QE$ is a height in $\Delta QBM$, and the heights $QE$, $BP$, $MS$ are concurrent. - $CP$, $AM$, $EL$ are the heights in $\Delta AEC$. Let $D$ be their intersection, the orthocenter of this triangle. We show that $A,Q,P,D,L$ are on a circle. First, $ALDP$ cyclic because of the right angles in $L,P$. Then the blue angle in $Q$ is equal to the blue angle in $L$ (because of the equality of each one with the blue angle in $E$), so $AQPL$ cyclic. In particular $\widehat{PLD}=\widehat{PAD}=y$. (Or we can extract this from $(PECL)$.) - The power of $A$ computed w.r.t. two circles gives the equality: $$ AL\cdot AC= AP\cdot AE = AQ\cdot AB\ . $$But $AB=AC$, so $AQ=AL$ and thus $QL\|BC$. - By the homothety / similitude transformation centered in $A$ which maps $\Delta ABC$ in $\Delta AQL$ the point $P\in AE$ corresponds to the point $R\in AE$ because $BP\|QR$. This gives then $CP\| LR$, and $LR\perp AE$, and $\widehat {LRQ}=\widehat {PCB}=y$. First proof: The idea is to use metric relations. (The following trigonometrical relations can be rephrased using similar triangles. I am not doing this, since the second proof will be synthetical.) - All the above steps are dancing around the problem, but at some point we need to introduce seriously $R$ and either the connection via $C$ using angles, or using metric relations. I decided to do the last. We will need the angle $\widehat{EQP} = \widehat{EQM} = \widehat{QMA} = 90^\circ-\widehat{PMB} = \widehat{PBM} = \hat B-y =90^\circ-\frac A2-y =90^\circ-\left(\frac A2+y\right)$. The sine theorem gives using e.g. $QP$ as an intermediate $$ \begin{aligned} \frac{PR}{PE} &= \frac{PR}{PQ} \cdot \frac{PQ}{PE} \cdot \frac{AC^2}{AC^2} = \frac{1}{\sin\widehat{QRP}} \cdot \frac{\sin\widehat{PEQ}}{\sin\widehat{EQP}} \cdot \frac{AC^2}{AC^2} \\ &= \frac{1}{\sin(A/2)} \cdot \frac{\sin y}{\cos\left(\frac A2+y\right)} \cdot \frac{AC^2}{AC^2} \\ &= \frac{AC^2\sin y}{MC\cdot AP} = \frac{AC^2}{MC} \cdot \frac{\sin \widehat {BPA}}{AB} = \frac{AC\sin(A/2)}{MC} \\ &=1\ . \end{aligned} $$- So $PR=PE$, and the height $CP$ in the triangle $\Delta CRE$ is also the median, so this triangle is isosceles, so we can finally also draw a green angle of measure $y$ in $\widehat {RCP}$. $\square$ Second proof: The idea is that the whole figure has $AM$ as a symmetry, so there correspond $A\leftrightarrow A$, $B\leftrightarrow C$, $Q\leftrightarrow L$. And the question is, what is corresponding to $D$? Well, $D$ is on the symmetry axis, so $D\leftrightarrow D$, so $D$ is the mid point of $EL$. (Draw the rectangle $EQL?$ if this is not a one second blitz.) So $D$ is the center of the circle $(EQRL)$, in particular, from $\color{blue}{DE}=DQ=\color{blue}{DR}=DL$ we need only the equality of the lengths of the) blue segments to get $\Delta DRE$ isosceles. So $PR=PE$, and then also $\Delta CRE$ isosceles. $\square$

First, let $AM$ intersect $(BRC)$ for the first and second time at $A'$ and $M'$. $\angle RAC+\angle RCA=\angle RBC'+\angle CBM'-\angle C'BA'=\angle RBC'+\angle CBA'-\angle C'BA'=\angle RBC'+\angle C'BC=\angle B$, so $\angle ARC=\pi-\angle RAC-\angle RCA=\pi-\angle B$. Let $H$ be the orthocenter of $ABC$. Then, $CHRA$ is cyclic, and since $CMPA$ is cyclic, $$\angle BCP=\angle MCP=\angle PAM=\angle RAM=\angle RCH=\angle RCP$$

Since $ACMP$ is cyclic, we have $\angle RPQ = \angle ACM$. Moreover, $\angle PQR = 90^\circ = \angle CMA$. Hence $\triangle PQR \sim \triangle CMA$. It follows that \[\frac{PR}{PQ} = \frac{CA}{CM}. \tag{1}\] Similarly, $\angle PMB = \angle PAC$ and $\angle BPM = 90^\circ = \angle CPA$. Hence $\triangle BPM \sim \triangle CPA$. It follows that \[ \frac{CA}{BM} = \frac{CP}{BP}. \tag{2} \] Using $(1)$, $(2)$, and the equality $BM=CM$ we obtain \[ \frac{PR}{PQ} = \frac{CA}{CM} = \frac{CA}{BM} = \frac{CP}{BP}. \tag{3}\] Using $(3)$ and $\angle QPB = 90^\circ = \angle RPC$ we obtain $\triangle QPB \sim \triangle RPC$. In particular, $\angle RCP = \angle PBQ$. Hence \[\angle RCP = \angle PBQ = \angle CBA - \angle MBP = \angle ACB - \angle ACP = \angle PCB.\]

Clearly, $\triangle MPB\sim\triangle APC~(\clubsuit)$ and that $\triangle RQP\sim\triangle AMC~(\spadesuit)$ and $$\angle RPC=\angle QPB=90^\circ$$Now $$\frac{BP}{PC}\overset{\clubsuit}{=}\frac{BM}{AC}=\frac{MC}{AC}\overset{\spadesuit}{=}\frac{QP}{RP}$$Consequently, $\triangle QPB\sim\triangle RPC$. Hence $$\angle RCP=\angle QBP=\angle ABP=\angle PAM=\angle PCM$$as desired. $\blacksquare$