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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?

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

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

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