BCQ = FCP

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BCQ = FCP

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#1 h Sep 1, 2023, 3:55 PM

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Let $\Gamma$ be the circumcircle of triangle $ABC (\overline{AB} > \overline{AC})$ . The bisector of $\angle BAC$ intersects $\Gamma$ at $D (\neq A)$ , and $E$ is the point on the segment $AB$ such that $\overline{AC} = \overline{AE}$ . Line $DE$ meets $\Gamma$ at $F (\neq D)$ . A line passing $E$ parallel to $BF$ intersects segments $AF$ , $BD$ at $P, Q$ , respectively. Prove that $\angle BCQ = \angle FCP$ .

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CyclicISLscelesTrapezoid

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#2 h Sep 1, 2023, 7:31 PM

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Notice that $DC=DB=DE$ , so $ACDE$ is a kite, which means $\overline{AD}$ bisects $\angle CDE$ . Thus, we have $AC=AF$ . We also have $DB=DC$ , so $E$ is the incenter of $\triangle CBF$ . We can restate the problem as follows.

Restated problem wrote:

Let $E$ be the incenter of $\triangle CBF$ , and let $D$ and $A$ be the midpoints of arcs $CB$ and $CF$ in the circumcircle of $CBF$ opposite $F$ and $B$ , respectively. Let the line through $E$ parallel to $\overline{BF}$ intersect $\overline{BD}$ at $Q$ and $\overline{FA}$ at $P$ . Prove that $\overline{CE}$ bisects $\angle QCP$ .

Consider an inversion at $E$ fixing the circumcircle of $CBF$ . This inversion maps $B$ to $A$ , $D$ to $F$ , and $C$ to the midpoint $M$ of arc $BF$ opposite $C$ . It also maps $P$ and $Q$ to the second intersections of the line through $E$ parallel to $\overline{BF}$ with the circumcircles of $BED$ and $FEA$ , which we call $P^*$ and $Q^*$ , respectively. It suffices to prove $\angle MP^*E=\angle MQ^*E$ . We have $\angle BP^*E=\angle BDE=\angle FAE=\angle FQ^*E$ , so $BFQ^*P^*$ is an isosceles trapezoid. The statement $\angle MP^*E=\angle MQ^*E$ follows by symmetry. $\square$

This post has been edited 3 times. Last edited by CyclicISLscelesTrapezoid, Sep 6, 2023, 11:20 PM

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#3 h Sep 3, 2023, 3:54 PM

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Nice. I have dif. solution, let's solve. Let $DF\cap BC=R$ . We know that $\angle RBQ=\angle CAD=\angle DAB=\angle DFB=\angle REQ$ which means that $(R E B Q)$ is cyclic. Then we have $\angle ACF=\angle ABF=\angle BEQ=\angle BRQ$ . Then $\angle ACF=\angle ACP+\angle PCF=\angle RQC+\angle RCQ$ , then if I will prove that $\sin(\angle ACP)/\sin(PCF)=\sin(\angle RQC)/\sin(\angle RCQ)$ , I will have $\angle PCF=\angle RCQ$ , as desired. Then let's prove it. Claim: $\sin(\angle ACP)/\sin(PCF)=\sin(\angle RQC)/\sin(\angle RCQ)$ By ratio lemma on $ACF$ with respect to $CP$ we have $\sin(\angle ACP)/\sin(PCF)=(AP/PF)*(CF/AC)$ $(1)$ .Since $PE//FB$ we have $AP/PF=AE/EB$ $(2)$ . We know that $\angle RFB=\angle REQ=\angle RBQ$ and $\angle BRQ=\angle BEQ=\angle EBF$ , which gives that $\triangle BEF\sim \triangle RQB$ , then $BE=RQ*(BF/BR)$ $(3)$ . Since $FR$ is angle bisector of $\angle CFB$ we have $BF/BR=CF/CR$ plugging it on $(3)$ gives that $BE=RQ*(CF/CR)$ , and plugging it on $(2)$ gives $AP/PF=(CR*AE)/(RQ*CF)$ , and plugging it on $(1)$ gives $\sin(\angle ACP)/\sin(PCF)=(AP/PF)*(CF/AC)=(AE/AC)*(CR/RQ)=CR/RQ$ and we know that $CR/RQ=\sin(\angle RQC)/\sin(\angle RCQ)$ , which proves the claim. So we are done :-D

:-D

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This post has been edited 3 times. Last edited by X.Allaberdiyev, Sep 3, 2023, 4:10 PM
Reason: Typo.

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CyclicISLscelesTrapezoid

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#4 h Sep 6, 2023, 11:32 PM

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Here's another solution to the restated problem in my above post (with points renamed for convenience).

Restated problem wrote:

Let $I$ be the incenter of $\triangle ABC$ , and let $\overline{BI}$ and $\overline{CI}$ intersect the circumcircle of $ABC$ again at $M_B$ and $M_C$ . Let the line through $I$ parallel to $\overline{BC}$ intersect $\overline{BM_C}$ and $\overline{CM_B}$ at $P$ and $Q$ , respectively. Prove that $\overline{AI}$ bisects $\angle PAQ$ .

Let $P'$ be the inverse of $P$ with respect to the circle centered at $M_C$ through $A$ , $B$ , and $I$ , and let $Q'$ be the inverse of $Q$ with respect to the circle centered at $M_B$ through $A$ , $C$ , and $I$ . We have $\angle BP'I=\angle M_CIP=\angle ICB$ and $\angle CQ'I=\angle M_BIQ=\angle IBC$ , so $P'$ and $Q'$ lie on the circumcircle of $BCI$ . We have $\angle IBP'=180^\circ-\angle IBM_C=180^\circ-\angle ICM_B=\angle ICQ'$ , so $IP'=IQ'$ . The circumcircle of $BCI$ is symmetric about $\overline{AI}$ by the incenter-excenter lemma, so $P'$ and $Q'$ are symmetric about $\overline{AI}$ . Therefore, we have $\angle BAP=\angle BAM_C-\angle M_CAP=\angle BCI-\angle AP'M_C=\angle IP'B-\angle AP'B=\angle AP'I=\angle AQ'I$ and analogously $\angle CAP=\angle AP'I=\angle AQ'I$ . Thus, $\overline{AP}$ and $\overline{AQ}$ are isogonal with respect to $\angle BAC$ , as desired.

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#5 h Mar 24, 2025, 10:27 AM

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Restated problem wrote:

Let $I$ be the incenter of $\triangle ABC$ , and let $\overline{BI}$ and $\overline{CI}$ intersect the circumcircle of $ABC$ again at $M_B$ and $M_C$ . Let the line through $I$ parallel to $\overline{BC}$ intersect $\overline{BM_C}$ and $\overline{CM_B}$ at $P$ and $Q$ , respectively. Prove that $\overline{AI}$ bisects $\angle PAQ$ .

Let $T$ denote the $A$ -mixtilinear incircle touchpoint, Let $O$ be the center of $(AIT)$ .
Claim 1: $O \in M_BM_C$ and $IO \parallel BC$
Proof: $O$ lies on perpendicular bisector of $AI$ , which is $M_BM_C$ . Notice that by reims we get that if $TI \cap (ABC)=N$ then the tangent at $I$ to $(AIT)$ is parallel to $NM_A$ , which is perpendicular to $BC$ but tangent at $I$ is perpendicular to $OI$ therefore $OI$ must be parallel to $BC$ . $\blacksquare$

Claim 2: $O = AA \cap TT$
Proof The tangency at $A$ follows by homothety at $A$ , the other tangency now follows as $OA=OT$

Notice by reims $M_BM_CPQ$ is cyclic, and so by power of point $OI^2=OA^2= OM_B \cdot OM_C = OP \cdot OQ$ which is enough to imply the isogonality, as desired.

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