Part (a) is trivial, just $\angle AVB=180-\angle ACB$. Let $N$ be the midpoint of $VO$, which is also the nine-point center. We show that $EP$ passes through $N$. Let $M$ be the midpoint of $VC$, then the nine-point circle also pass through $M$. So $\angle NPC=\angle NMV=\angle DCP$. But since $\angle CPD$ is right, $PN$ must pass through the midpoint $E$ of $CD$, and done.

A simpler solution.. Let $CO$ meets the circle at $C'$ and $M=PE\cap VO$. Since the triangles $\bigtriangleup CPD$ and $\bigtriangleup CV'C'$ are right-angled and $E,D$ are the midpoints of $CD$ and $CC'$ respectively, we easily get that $PE \parallel V'O$. Therefore $\frac{VM}{MO}=\frac{VP}{PV'}=1$

mhet49 wrote: Let $ABC$ be a triangle where $AC\neq BC$. Let $P$ be the foot of the altitude taken from $C$ to $AB$; and let $V$ be the orthocentre, $O$ the circumcentre of $ABC$, and $D$ the point of intersection between the radius $OC$ and the side $AB$. The midpoint of $CD$ is $E$. a) Prove that the reflection $V'$ of $V$ in $AB$ is on the circumcircle of the triangle $ABC$. b) In what ratio does the segment $EP$ divide the segment $OV$? Another way to solve part b) Let $K$ intersection of $VO$ and $PE$ than from menelaus theorem we have $1=\frac{PV\cdot CE\cdot OK}{PC\cdot EO\cdot KV}$ than from $PV=PV'$ and from $CPO$ is similar with$COV'$ so we have $OK=KV$.

(a) We have that $\widehat{AVP}=\frac \pi 2-\widehat B$ and $\widehat{BVP}=\frac \pi 2-\widehat A$, so that $\widehat {AVB}=\pi-\widehat C$, from which the statement follows ($CAV'B$ is cyclic). (b) An homotethy centered at $C$ shows that $OV' \parallel EP$; then by Thales theorem, to prove that $EP$ bisects $VO$ it is enough to prove that $P$ bisects $VV'$, but this is actually the result from point (a)!

a)Extend $BV$ to $BF$ where $F$ lies on $AC$. Extend $AV$ to $D$ such that $D$ lies on $BC$. As, $V'$ is the reflection of V on $BP$ so, $BV=BV'$. So,$\angle BV'V=\angle BVV' =\angle BDP=90^{\circ}- \angle ADP=90^{\circ}-\angle ACV'$..............(1) Again, $\angle BV'V=90^{\circ}-\angle ABV'$......................(2) from (1) and (2) we get $90-\angle ACV'=90^{\circ}-\angle ABV'$ . SO, $\angle ACV=\angle ABV'$ So, $V'$ lies on the circumcircle of triangle $ABC$

(a) is well known. (b) Let $PE \cap VO = X$. By (a), it's enough to show that $VPX \sim VV'O$ or $PE || V'O$ which is true because $\triangle EPC \sim \triangle OV'C$.

a) it's enough to show ∠AVT + ∠C = 180. ∠AVT = 180 - ∠BAV + ∠ABV = 180 - ∠C. b) Let VO and PE meet at S. ∠CV'O = ∠V'CO and ∠CPE = ∠ECP ---> V'O || PE ---> VP/VV' = VS/VO ---> VS/VO = 1/2

Part A: This is trivialized by the Orthocenter Reflection Lemma. Part B: Let $C_1$ be the $C$-antipode wrt $(ABC)$, the midpoint of $AB$ be $X$, and the midpoint of $CV$ be $Y$. It's well-known that $C_1$ and $V$ are symmetric about $X$, so properties of midlines yield $XY \parallel CC_1$. Now, taking a homothety at $P$ implies $PE$ bisects $XY$, i.e. it meets $XY$ at the Nine-Point center $N_9$. But it's well-known that $N_9$ is the midpoint of $OV$, so our answer is $1$. $\blacksquare$

1:51.05 a) trivial, just have to show $\angle AVB = 180 - \angle A$ which is very easy. b) $\measuredangle CV'O = \measuredangle V'OC = \measuredangle CPE$ since $E$ circumcenter of $CPD$ hence $\overline{EP} \parallel \overline{OV'}$, but if $\overline{EP} \cap \overline{OV} = X$ then by homothety $XP$ is the midline in triangle $VV'O$. answer is $1$, done.

Part a) is trivial: $\measuredangle AV'C = \measuredangle V'VA = \measuredangle ABC. \blacksquare$ For part b), consider angles modulo $\pi$. I claim that $X$ is the nine-point center (midpoint of $OV$) of $(ABC)$. We make the following lemma: Claim: $EP \parallel OV'$. Proof: Observe that \[\measuredangle CV'O = \measuredangle OCP = \measuredangle ECP = \measuredangle CPE. \square\]Now as $P$ is the midpoint of $\overline{VV'}$, then $\overline{XP}$ must be the midline of $\triangle VOV'$, so that $X$ is the midpoint of $VO$. Therefore $\boxed{VX/XO = 1}$, as desired. $\blacksquare$