Cyclic intersections of $AH$ and $BO$ gives similar $\Delta$

[rah4927]

In $\triangle ABC$, let $H$ and $O$ be the orthocentre and circumcentre respectively. Now, let $X=AH\cap BO,Y=BH\cap CO, Z=CH\cap AO$. Now, prove that, $\triangle ABC$ and $\triangle XYZ$ are similar.

[Thanic Nur Samin]

We use complex numbers. Use the lowercase letters to denote the coordinate of a point denoted with higher case form of the same letter. Take the circumcircle of $\Delta ABC$ as the unit circle.

Now, let $AH$ intersect the circumcircle at $P_1$ and $BO$ intersect the circumcircle at $Q_1$. Now, the coordinates of $P_1$ and $Q_1$ are respectively $-\dfrac{bc}{a}$ and $-b$. So, their intersection point X has coordinate $x=\dfrac{b(a^2-bc)}{a(b-c)}$. Determine $y$ and $z$ similarly. Now, $\Delta ABC$ and $\Delta XYZ$ will be similar oppositely if and only if $\begin{vmatrix} \bar{a} & x & 1 \\ \bar{b} & y & 1\\ \bar{c} & z & 1 \end{vmatrix}=0$ or $\begin{vmatrix} \dfrac {1} {a}& \dfrac {b\left( a^{2}-bc\right) } {a\left( b-c\right) }& 1\\ \dfrac {1} {b}& \dfrac {c\left( b^{2}-ca\right) } {b\left( c-a\right) }& 1\\ \dfrac {1} {c}& \dfrac {a\left( c^{2}-ab\right) } {c\left( a-b\right) }&1 \end{vmatrix}=0$, which is true.

So $\Delta ABC$ and $\Delta XYZ$ are oppositely similar.

[joydip]

$\angle XAC =\angle OAB=\angle ABX$ .so $\odot AXB$ touches $AC$ at $A$ .Similarly $\odot BYC$, $\odot AZC$ touches $AB$ & $BC$ at $B$ & $C$ respectively .By Miquel's theorem $\odot AXB$ , $\odot BYC$ & $\odot AZC$ pass though a common point, say $P$ .(Using directed angles) $ \angle XPY = \angle XPB + \angle BPY $. $ \angle XPB= \angle XAB = 90^{\circ} -\angle B$ ,$\angle BPY= \angle BCY =90^{\circ} -\angle A$.so, $ \angle XPY =180 - \angle A - \angle B = \angle C =\angle XHY$ .So $HYPX$ is cyclic. Similarly $HZPX$ is cyclic.So $H,P,X,Z,Y$ are concyclic.$\angle XYZ=\angle XHZ=\angle B$ and others similerly.So $\triangle ABC$ and $\triangle XYZ$ are similar.