# Point $D$ is on line $BO$ such that $O$ is between $B$ and $D$ and $\angle ADC = \angle ABC$.

There is a $$\Delta ABC$$ with $$AB < BC$$ , with orthocenter $$H$$ and circumcenter $$O$$. Point $$D$$ is on line $$BO$$ such that $$O$$ is between $$B$$ and $$D$$ and $$\angle ADC = \angle ABC$$. The semi-line starting at $$H$$ and parallel to $$BO$$ meets the circumscribed circle at points $$E$$ and $$F$$ respectively. Prove that $$BH = DE$$.

What I Tried: Here's a picture in Geogebra to get an idea of the problem :-

Really Tough. Spent around $$1.5$$ hours on this problem but no progress, as I am a little weak at Geometry I tried elementary methods like angle-chasing , similarity and so on, but no success. Another problem is that $$DABC$$ is not a parallelogram as it may seem, so angle-chasing and similarity do not work to any triangles at all till now, so that I could have found something useful.

We basically need to prove that $$DEHB$$ is an Isosceles Trapezium, and I have no idea. I did not even any reasonable cyclic quadrilaterals to work on and gain some information. I am quite stuck on this problem.

Can anyone help me?

Let $$O', H'$$ be the circumcenter and the orthocenter of $$\triangle ADC$$ respectively. Denote, furthermore, by $$G$$ the orthogonal projection of $$O$$ onto $$CA$$.

First of all, observe that the Law of Sines implies that the circumcircle of $$\triangle ADC$$ has the same radius as the circumcircle of $$\triangle ABC$$, since $$R'=\frac{CA}{\sin\angle CDA}=\frac{CA}{\sin\angle ABC}=R$$. It follows that $$G$$ is the midpoint of $$OO'$$.

Besides, it is well-known (at least considering the ratios of the Euler line), that $$2OG=BH\implies OO'=BH$$ and since both $$OO'$$ and $$BH$$ are perpendicular to $$CA$$, we have that $$OO'HB$$ is a parallelogram. Yet this implies that $$OB\parallel O'H$$, and, hence, $$O',E,H$$ are collinear. Notice now that, since $$OE'\parallel OD$$ and $$DO'=R'=R=EO$$, the quadrilateral $$DO'EO$$ is an isosceles trapezium; we infer that the diagonals have the same length, i.e. $$DE=OO'$$. Thus $$DE=OO'=BH$$

Some additional observations. As I noticed while playing with Geogebra, this problem could lead to other nice relationships such as

• The Nine-point-Center $$N_9'$$ of $$\triangle ADC$$ lies on $$BD$$.
• $$H'$$ and $$E$$ are symmetric wrt $$BD$$.
• Nice answer. As you've observed reflection is the key; it introduces several parallelograms into the figure. Dec 28, 2020 at 11:26
• @cosmo5 Considering the parallelograms was, in fact, key. It always surprises me that these solutions follow simply from $2OG=BH$ and some parallelograms... Dec 28, 2020 at 11:32
• I don't know the reason, but they seem dual/complementary to the sides in a way, hence quite important. $a=2R\sin A$, $AH=2R\cos A$, $a^2+AH^2=(2R)^2$. Dec 28, 2020 at 11:35
• Great Solution, +1 . I never thought you would have got such great observations while working with reflections. Dec 28, 2020 at 12:23