AHZOLFAGHARI wrote: Consider the triangle $ABC$ . The points $D,E$ are on sides $AB,AC$ such that $BDEC$ is cycle . The common point of $BE$ and $CD$ is $P$ . The point $ H$ is on $AC$ such that $\angle PHA = 90 $ . If $M,N$ in the midpoint of $AP,BC$ prove that : $ AC\color{red}{ D }\normalcolor \sim MNH $ . Typo corrected My solution: Let $ F, G $ be the midpoint of $ CP, CA $, respectively . Since $ F, G, H, M $ are concyclic ( 9 point circle of $ \triangle APC $ ) , so combine $ \angle FMG=\angle ECD=\angle EBD=\angle FNG \Longrightarrow F, G, H, M, N $ are concyclic , hence from $ \angle ACD=\angle HGM=\angle HNM , \angle MHN=\angle MFN=\angle AEB=\angle ADC \Longrightarrow \triangle ACD \sim \triangle MNH $ . Q.E.D

TelvCohl wrote: AHZOLFAGHARI wrote: Consider the triangle $ABC$ . The points $D,E$ are on sides $AB,AC$ such that $BDEC$ is cycle . The common point of $BE$ and $CD$ is $P$ . The point $ H$ is on $AC$ such that $\angle PHA = 90 $ . If $M,N$ in the midpoint of $AP,BC$ prove that : $ AC\color{red}{ D }\normalcolor \sim MNH $ . Typo corrected My solution: Let $ F, G $ be the midpoint of $ CP, CA $, respectively . Since $ F, G, H, M $ are concyclic ( 9 point circle of $ \triangle APC $ ) , so combine $ \angle FMG=\angle ECD=\angle EBD=\angle FNG \Longrightarrow F, G, H, M, N $ are concyclic , hence from $ \angle ACD=\angle HGM=\angle HNM , \angle MHN=\angle MFN=\angle AEB=\angle ADC \Longrightarrow \triangle ACD \sim \triangle MNH $ . Q.E.D Yes , thanks , edited .

Another (more complicated idea): Let $K$ the projection of $P$ onto $AB$, $L, Q$ the midpoints of $DE, KH$. Known properties: $M,L,N$ determine the Newton-Gauss line of $ADPE$ and $Q$ belongs to this line. Also $KN=NH$, easy to prove: take $O', O"$ the midpoints of $BP, PC$ and see congruent triangles, so $MN\bot KH$. Next, use the following lemma: If, on the sides $KP, PH$ of a triangle $KPH$ are constructed the similar triangles $BPK, CPH$ and $X$ is the intersection of the perpendicular bisectors of $KH, BC$ respectively, then $m(\widehat{KXH})=2m(\widehat{KBP})$ Proof Let $O',O"$ the circumcenters of the triangles $BPK, CPH$ and $DYH$ an isosceles triangle similar to $BPK, CPH$, with $P$ and $Y$ on the same side of $KH$. Known property: $PO'YO"$ is a parallelogram (a spiral similarity kills the problem), and $\angle DYH=\angle PO"H$ Likewise, for $\triangle BPC$ with same similar triangles $BPK, CPH$ we have $\triangle BO'P\sim\triangle PO"C$ (isosceles) and construct the isosceles triangle $BZC$ similar to $BO'P$, with P and Z on the same side of $BC$. By same property, $PO'ZO"$ is a parallelogram, hence $Z\equiv Y\equiv X$. Now to problem: $X\equiv N$ and $\angle DNH=2\angle DCA$. Also $M$ is the circumcenter of $AKPH$, so $\angle DMH=2\angle BAC$, so $MKNH$ is a kite with vertices angles $M, N$ being double of $\angle BAC$ and $\angle ACD$, hence we are done. Best regards, sunken rock

It's easy from nine-point circle and midline ! Nice problem from Iran !

My first solution: Let $R$ midpoint of $DE$ and $H'$ be the projection of $P$ on $AB$ points $M,N,R$ lie on newton gauss line of quadrilateral $ADPE$ according to this problem we deduce that $RN\perp HH'$ because $MH=MH'$ we get that $MN$ is perpendicular bisector of $HH'$ note that $M$ is center of cyclic quadrilateral $AHPH'$ so $\angle HMH'=2\angle A\longrightarrow \angle NMH=\angle A$ from IMO shotlist G4 2009 $MP$ is tangent to $\odot (\triangle RPN)$ so $\triangle MPR\sim \triangle MNP\longrightarrow \frac{MN}{MP}=\frac{PN}{PR}$ because $MP=MH\longrightarrow \frac{MN}{MH}=\frac{PN}{PR}$(1) but $\triangle PED\sim \triangle PCB\longrightarrow \frac{PN}{PR}=\frac{PE}{PC}=\frac{AD}{AC}$(2) from (1),(2): $\frac{AD}{AC}=\frac{MN}{MH}\longrightarrow \triangle MNH\sim \triangle DCA$ DONE

My second solution: $S,T$ are reflections of $P$ throw $H$ and $N$ clearly $\triangle ASC=\triangle APC$ because $HN|| ST,HM|| SA\longrightarrow \triangle MHN\sim \triangle AST$ from IMO shortlist G2 2012 quadrilateral $ASCT$ is cyclic $\longrightarrow \angle ATS=\angle ACS=\angle ACD$(1) quadrilateral $PCTB$ is parallelgram so $EB|| CT\longrightarrow \angle ADC=\angle AEB=\angle ACT=\angle AST$(2) from (1),(2) $\triangle ACS\sim \triangle ADC\sim \triangle MNH$ DONE

My solution: Lemma. Given $\triangle{ABC} \sim \triangle{DEF}$ (same direction) $X, Y, Z\in{AD, BE, CF}: \frac{XA}{XD} = \frac{YB}{YE} = \frac{ZC}{ZF}$ We will have: $\triangle{ABC} \sim \triangle{DEF} \sim \triangle{XYZ}$ Proof. Vector rotating. Back to our main problem. $X, Y, Z, S$ are the midpoints of $DC, CA, AD, DE$ $\Rightarrow \overline{M, N, S}, \overline{Y, M, Z}, \overline{Z, S, X}, \overline{X, N, Y}$ $Q$ is the reflection of $P$ WRT $CE$ We have: $\triangle{PDB} \sim \triangle{PEC} \Rightarrow \triangle{PDB} \sim \triangle{QEC}$ (same direction) But $H, S, N$ are the midpoints of $PQ, DE, BC$ $\Rightarrow \triangle{PDB} \sim \triangle{QEC} \sim \triangle{HSN}$ (lemma) $\Rightarrow \triangle{HSN} \sim \triangle{PEC}$ $\Rightarrow \frac{NH}{NS} = \frac{CP}{CE}$ (1) On the other hand: $\frac{NS}{NM} = \frac{XS}{XZ}.\frac{YZ}{YM} = \frac{CE}{CA}.\frac{CD}{CP}$ (2) (1), (2) $\Rightarrow \frac{NH}{NS}.\frac{NS}{NM} = \frac{CP}{CE}.\left(\frac{CE}{CA}.\frac{CD}{CP}\right)$ $\Rightarrow \frac{NH}{NM} = \frac{CD}{CA}$ $\Rightarrow \triangle{HNM} \sim \triangle{DCA}$ (s.a.s) Q.E.D

TelvCohl wrote: AHZOLFAGHARI wrote: Consider the triangle $ABC$ . The points $D,E$ are on sides $AB,AC$ such that $BDEC$ is cycle . The common point of $BE$ and $CD$ is $P$ . The point $ H$ is on $AC$ such that $\angle PHA = 90 $ . If $M,N$ in the midpoint of $AP,BC$ prove that : $ AC\color{red}{ D }\normalcolor \sim MNH $ . Typo corrected My solution: Let $ F, G $ be the midpoint of $ CP, CA $, respectively . Since $ F, G, H, M $ are concyclic ( 9 point circle of $ \triangle APC $ ) , so combine $ \angle FMG=\angle ECD=\angle EBD=\angle FNG \Longrightarrow F, G, H, M, N $ are concyclic , hence from $ \angle ACD=\angle HGM=\angle HNM , \angle MHN=\angle MFN=\angle AEB=\angle ADC \Longrightarrow \triangle ACD \sim \triangle MNH $ . Q.E.D How did you get $\angle HGM=\angle ACD$ and $\angle MFN=\angle AEB$? And also, how did you come up with this solution?

Let $K,L$ be the feet of $P$ on $AB,BC \ . \ MH=MK= \frac{AP}{2},\widehat{KMH}=2 \ \hat A ,HKNL $ are on the circle of the feet of $P$ and its isogonal thus $\widehat{HKN}=\widehat{HLN}=\widehat{HPC}=\widehat{BPK}=\widehat{BLK}=\widehat{NHK}$ hence $ NH=NK$ so $MHN$ is congruent to $MKN$ then it suffices to prove that $\widehat{KNH}=2 \cdot \widehat{DCA} $ but $ \widehat{KNH}=\widehat{KNP}+\widehat{PNH}=\widehat{KBP}+\widehat{PCH}=2 \cdot \widehat{DCA}$ . R HAS

1. Let $E'$ be reflection of $E$ over $H$. 2. Triangles $ABE$ and $PCE'$ are similar by AA Similarity. They have same orientation. 3. Triangle $ABE$ and $MNH$ are similar by Mean Geometry / Spiral similarity theory, 4. $ACD ~ ABE$ by AA similarity and thus $ACD ~ MNH$ by transitive property.

TelvCohl wrote: AHZOLFAGHARI wrote: Consider the triangle $ABC$ . The points $D,E$ are on sides $AB,AC$ such that $BDEC$ is cycle . The common point of $BE$ and $CD$ is $P$ . The point $ H$ is on $AC$ such that $\angle PHA = 90 $ . If $M,N$ in the midpoint of $AP,BC$ prove that : $ AC\color{red}{ D }\normalcolor \sim MNH $ . Typo corrected My solution: Let $ F, G $ be the midpoint of $ CP, CA $, respectively . Since $ F, G, H, M $ are concyclic ( 9 point circle of $ \triangle APC $ ) , so combine $ \angle FMG=\angle ECD=\angle EBD=\angle FNG \Longrightarrow F, G, H, M, N $ are concyclic , hence from $ \angle ACD=\angle HGM=\angle HNM , \angle MHN=\angle MFN=\angle AEB=\angle ADC \Longrightarrow \triangle ACD \sim \triangle MNH $ . Q.E.D I think $ \angle ACD=\angle HGM=\angle HNM$ not right... in part $\angle HGM$, which has contradiction with the case when $G$ is closer than $H$ to $A$ and would rather be $ \angle ACD=\pi - \angle HGM(=\angle AGM)=\angle HNM$ in this case

great problem to work on

Let $R$ be the reflection of $P$ over $N$ and $S$ be the reflection of $P$ over $H$. As $PH \perp AC$, reflection over $H$ is same as reflection over $AC$. Lemma 1: $AR$ and $AP$ are isogonal wrt $\angle BAC$. Proof: In $\triangle ADP$ and $\triangle ACR$, $\angle ADP=\angle ADE+\angle PDE=\angle ADE+\angle PBC=\angle ACB+\angle BCR=\angle ACR$. [by reflection we had that $PBRC$ as a parallelogram.] And, $\frac{DP}{CR}=\frac{DP}{PB}=\frac{DE}{BC}=\frac{AD}{AC}$. So, $\triangle ADP \sim \triangle ACR$. or, $\angle DAP= \angle CAR \Rightarrow \angle DAR +\angle RAP = \angle CAP+\angle PAR \Rightarrow \angle DAR=\angle CAP$. Which implies $AR$ and $AP$ are isogonal wrt $\angle BAC$. Lemma 2: $A, R, C, S$ are concyclic. Proof: $180^o-\angle ASC =\angle SAC+\angle SCA=\angle PAC+\angle PCA=\angle APD$. And we've shown that $\triangle ADP \sim \triangle ACR$, which implies $\angle APD=\angle ARC$. So, $\angle ARC= 180^o-\angle ASC \Rightarrow \angle ARC+\angle ASC=180^o$. So, $A, R, C, S$ are concyclic. Lemma 3: $\triangle ADC \sim \triangle ASR$. Proof: We've shown that $\angle ADC=\angle ACR$. and from lemma 2 $\angle ACR=\angle ASR$. So, $\angle ADC=\angle ASR$. And from lemma 1, $\angle DAR=\angle PAC=\angle CAS=a$ [last one is from reflection]. So, $\angle DAC=\angle DAR+\angle RAM+\angle PAC= 2a+\angle RAP$. And, $\angle SAR= \angle RAP+\angle PAC+ \angle CAS=\angle RAP+2a$. So, $\angle DAC= \angle SAR$. Which implies $\triangle ADC \sim \triangle ASR$. Lemma 4: $\triangle ASR \sim \triangle MHN$. Proof: $N,H,M$ are the midpoint of the side $\overline{PR}, \overline{PS}, \overline{PA}$. So, $MN \parallel AR, MH \parallel AS, HN \parallel SR$. So, $\triangle SAR \sim \triangle HMN$.[Note this cam be done with a homothety center at $P$ with ratio $-\frac{1}{2}$]. From lemma 3, $\triangle ADC \sim \triangle ASR$. From lemma 4 $\triangle ASR \sim \triangle MHN$. So, $\triangle ADC \sim \triangle MNH$. $\mathbb Q. \exists. \mathbb D.$

Nice one Let S,Q be midpoints of CP and CA. lemma1 : MHQSN is cyclic. proof: we will prove MHQS and HQSN are cyclic. ∠AHP = 90 ---> ∠AHM = ∠MAH = 180 - ∠APC - ∠ACP = 180 - ∠QSC - ∠MSP = ∠QSM ---> MHQS is cyclic. ∠PHC = 90 ---> ∠SHQ = ∠SCH = ∠PBD = ∠SNQ ---> HQSN is cyclic. ∠ACD = ∠AQM = ∠HNM and ∠DAC = ∠NQC = ∠NMH so NHM and CDA are similar.

suli wrote: 1. Let $E'$ be reflection of $E$ over $H$. 2. Triangles $ABE$ and $PCE'$ are similar by AA Similarity. They have same orientation. 3. Triangle $ABE$ and $MNH$ are similar by Mean Geometry / Spiral similarity theory, 4. $ACD ~ ABE$ by AA similarity and thus $ACD ~ MNH$ by transitive property. Would someone please explain step 2?