We have to prove MP=MR=1/2 AB Assume coordinate $A=(0,0),B=(2b,2c),c=(2a,0)$thus$M=(a+b,c),p=(a,2c)$ so PM=√$(b^2+c^2)$ , AB=2√$(b^2+c^2)$ so PM=MR Q.E.D

Hint: if $PQ$ meets $AB$ at $R$ show that $PCRB$ is a parallelogram.

WakeUp wrote: Hint: if $PQ$ meets $AC$ at $R$ show that $PCRB$ is a parallelogram. Typo edited. Now we have $AC//BP$ and $BM=MC$ hence $RMC,PMB$ are congruent triangles (ASA), so $RC=BP$ and $PCRB$ a parallelogram. But the dificulty was to prove $ARPB$ is an isoceles trapezoid. Why does $PM=MR$ solves the problem?

Let the base of the perpendicular bisector of segment AC be a point L. It is easy to get that BPCR is a parallelogram. Furthermore, since triangle LRP is a right triangle, so LM=RM=PM. But on the other hand M is a midpoint of BC,therefore 2LM=AB=RP, so we are done...

Let $R$ be midpoint of $AC\implies PR$ is the projection of $BC$ onto the perpendicular bisector of $AC$, so $M$ is equally apart from $P, R$. Now let $\{S\}=AB\cap PR$; with $MR\parallel AB\implies \triangle PQS\sim\triangle PMR$, so $\triangle PSQ$ is isosceles; with $BP\bot PS$, we are done. Best regards, sunken rock

Another solution : N , R is the midpoint of [AB] and [AC] using thales we are going to prove : QN = QM note that : $\angle {QNM} = A$ in other side we have : EPM isosceles (because if we note : L IS the proje. of M into (EP) we have : LP = LE) and so : $\angle {MEP} = \angle {MPE} = 90 - A$ But we have : $\angle {PMN} = 9O - \angle{EPM} = A$ and so : QNM is isosceles hence QM = QN !! we are done !!

Dear mathlinkers, I haven't read the posts you posted, but I want to post my solution, I hope it have some difference. Denote $K$ is the refection of P through $M, N$ is the midpoint $AC$. So $PCKB$ is a parallelogram $\Rightarrow CK \| BP$. But $BP \| AC$, so $K$ is on $AC$. It follows $M$ is the midpoint of side $PK$ of the triangle $PNK$ with $\angle{PNK}=90^0$. So $MN=MP \Rightarrow \angle{MNK}=\angle{QPB} \Rightarrow \angle{QBP}=\angle{QPB} \Rightarrow QP=QB$.

Proof. Let $N$ be midpoint of $BC$, from $BP\parallel AC$ and $MN\parallel AB$ we have \[-1=(ACN)=B(ACNP)=(RSPN)=M(RSPN)=M(RBQN)=(RBQ)\] So $Q$ is midpoint of $BR$ but $\angle BPR=90^\circ$ we have $QB=QP$. General problem. Let $BN$ be median of triangle $ABC$. $M$ is a point on $BC$. $S$ lies on $BN$ such that $MS\parallel AB$. $P$ is a point such that $SP\perp AC$ and $BP\parallel AC$. $MP$ cuts $AB$ at $Q$. Prove that $QB=QP$.

Proof.

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nsun48 wrote: Given triangle $ABC$. Point $M$ is the midpoint of side $BC$, and point $P$ is the projection of $B$ to the perpendicular bisector of segment $AC$. Line $PM$ meets $AB$ in point $Q$. Prove that triangle $QPB$ is isosceles. buratinogigle wrote: Proof. Let $N$ be midpoint of $BC$, from $BP\parallel AC$ and $MN\parallel AB$ we have \[-1=(ACN)=B(ACNP)=(RSPN)=M(RSPN)=M(RBQN)=(RBQ)\] So $Q$ is midpoint of $BR$ but $\angle BPR=90^\circ$ we have $QB=QP$. General problem. Let $BN$ be median of triangle $ABC$. $M$ is a point on $BC$. $S$ lies on $BN$ such that $MS\parallel AB$. $P$ is a point such that $SP\perp AC$ and $BP\parallel AC$. $MP$ cuts $AB$ at $Q$. Prove that $QB=QP$.

Proof.

I want prove that M(BRQN)=-1 but M(BRQN)=(SRPN)=B(SRPN)=(CAN)=-1.QED!

let $A\equiv(0,0)$ ,$C\equiv(a,0)$ ,and $B\equiv(b,c)$ so , we have by simple geometry that ,$P\equiv(\frac{a}{2},c)$ so , by finding equations of $AB$ and $PM$ ,we have , $Q\equiv(\frac{2b+a}{4},\frac{c(2b+a)}{4b})$ so , we have $QB^2=QP^2$. hence done!

buratinogigle wrote: Proof. Let $N$ be midpoint of $BC$, from $BP\parallel AC$ and $MN\parallel AB$ we have \[-1=(ACN)=B(ACNP)=(RSPN)=M(RSPN)=M(RBQN)=(RBQ)\] So $Q$ is midpoint of $BR$ but $\angle BPR=90^\circ$ we have $QB=QP$. General problem. Let $BN$ be median of triangle $ABC$. $M$ is a point on $BC$. $S$ lies on $BN$ such that $MS\parallel AB$. $P$ is a point such that $SP\perp AC$ and $BP\parallel AC$. $MP$ cuts $AB$ at $Q$. Prove that $QB=QP$.

Proof.

Can you prove it?

wow guys...synthetic solutions plz not coordinates Let MP intersect AC = R. Then since triangles BPM and triangles CMR are congruent (AAS), we have MP = MR. Let midpoint of AC be N. Then MP = MR = MN since triangle PNR is right. MN = 1/2AB by midline, so MP = 1/2AB. Thus, MR = AB and <ARM = <A = <BPQ by parallel lines. Finally, we have <QBP = <A by parallel lines as well, so triangle BPQ is isosceles.

Let O be circumcenter of triangle ABC. BPMO is cyclic quadrilateral so <MPO=<MBO=90 - <BAC, <BPO=90 thus <BPQ=<BAC and <PBQ=<BAC because BP || AC

Dear Mathlinkers, 1. T the symmetric of B wrt the perpendicular bissector of AC 2. ABTC is an isoceles trapeze 3. TC // PM... and we are done... Sincerely Jean-Louis

Since $AC$ and $BP$ are parallel, $\measuredangle ACB=\measuredangle PBC.$ Then, $\measuredangle RPB=\measuredangle QRA= \measuredangle PRC$, implying that $\triangle PMB \sim \triangle RMC$. This means that $RM=PM$, so $RCPB$ is a parallelogram. Hence $BR=PC$. Notice that $\triangle ACP$ is isosceles, so $BR=PC=PA$. Therefore, $ARPB$ is an isosceles trapezoid, which then immediately implies the result since $\measuredangle ABP=\measuredangle BPR$.

we denote the circumcenter of $\triangle{ABC}$ as $O$, Join $PM$ and $OM$ since we have $OM \perp BC$ and $BP \perp$ the perpendicular bisector , quadrilateral $BPMO$ is cyclic and since $\angle{POM}=\angle{ACB}$ we have $\angle{PMC}=\angle{ACB}$ hence $\angle{QBP}=\angle{BAC}$ also we have $\angle{OBM}=\angle{OPM}=90^{\circ}-\angle{BAC}$ we have $\angle{BPQ}=\angle{BAC}$ hence we have $\triangle{QPB}$ as isosceles triangle $\blacksquare$

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