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In an acute-angled triangle $ABC$ with $AB < AC$, the circle $\Gamma$ touches $B$ and intersects $AC$ at $D$ and then passes through $C$. Prove that the orthocentre of $\Delta ABD$ lies $\Gamma$ if and only if it lies on the perpendicular bisector of $\Delta ABC$.

I tried to first prove that if $H$ lies on the perpendicular bisector, $\Gamma$. Assume that for some $\Delta ABC$, $H$ lies on the perpendicular bisector. Consider the following figure. The obvious thing to prove was that $HDCB$ is cyclic.

Now, let $\angle FBH = \alpha, \angle HBD = \alpha$. As $AB$ is tangent to $\Gamma$, by the special case of the inscribed angle theorem, $\angle FBD = \angle BCD = \alpha + \beta$

Construct through $B$ a line parallel to $HD$. Let it meet the circle at $K$ (not shown in fig). If we assume that $HDCB$ is cyclic, then $\angle DHB = 180 - \alpha - \beta$. But this, in turn would mean that $\angle HBK = \alpha + beta \implies \angle DBK = \beta$. But this would mean that $\angle HDB = \beta$.

Now, in $\Delta FBH$, it is clear that $\angle FHB = 90 - \beta = \angle GHD$. This means $\angle ADF = \beta$. Now this would mean that $DF$ is an altitude as well as an angle bisector, which is only possible if $\Delta ABD$ is isosceles. So, the problem reduces to:

Prove that if $H$ lies on the perpendicular bisector, $HI$, $\Delta ABD$ is isosceles.

I couldn't prove this.

I also started over and resorted to angle-chasing, but could get nowhere. I couldn't think of any way to use the fact that $BI =IC$, and so couldn't use all the data.

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  • $\begingroup$ you give a wrong direction for the bisector. it is the bisector of AC as dtldarek said. Another approach is to prove $AH=CH$ and you will get answer at once. $\endgroup$
    – chenbai
    Dec 18, 2013 at 2:56

2 Answers 2

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Angle-chasing is helpful.

Let $Q$ be the orthocenter of $\triangle ABD$, the common point on altitudes $\overline{AA^\prime}$, $\overline{BB^\prime}$, $\overline{DD^\prime}$. Let $G$ be the point where $\overline{DD^\prime}$ meets the circle.

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Right triangles $\triangle ABA^\prime$ and $\triangle DBD^\prime$ overlap at $\angle B$, so their "other" acute angles match, with, say, measure $\theta$; likewise, the non-overlapped acute angles of $\triangle ADA^\prime$ and $\triangle BDB^\prime$, with measure $\phi$. In the circle, inscribed angles $\angle GDB^\prime$ and $\angle GBD^\prime$ subtend the same arc, $\stackrel{\frown}{BG}$, so that $|\angle GBD^\prime| = |\angle GDB| = \theta$.

Now the implication chain can begin:

$$\begin{align} Q \text{ lies on the circle } \quad &\Leftrightarrow \quad Q \text{ and } G \text{ coincide[*]} \\ &\Leftrightarrow \quad |\angle QBG| = 0 \\ &\Leftrightarrow \quad |\angle DBA| = |\angle DAB| = \theta + \phi \\ &\Leftrightarrow \quad \triangle ADB \text{ is isosceles with base } \overline{AB} \\ &\Leftrightarrow \quad \text{altitude } \overline{DD^\prime} \text{ is also a perp. bis. of } \overline{AB} \\ &\Leftrightarrow \quad Q \text{ lies on the perp. bis. of } \overline{AB} \end{align}$$

[*] Why can't $Q$ and $D$ coincide?

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  • $\begingroup$ The fact that $\Delta ABD$ is isosceles is enough to prove the statement, since I had already reduced the problem to that point. But, how does your conclusion that $Q$ lies on the perp. bis. of $AB$ help? $\endgroup$
    – Gerard
    Dec 27, 2013 at 5:53
  • $\begingroup$ The altitude to a side of a triangle coincides with the perpendicular bisector of that side when (and only when) the triangle is isosceles with the side as its base. (The altitude already has the "perpendicular" property; at question is whether it also has the "side-bisecting" property.) By construction, $Q$ is on the altitude to $\overline{AB}$; in (and only in) isosceles circumstances, $Q$ is also on the perpendicular bisector, since the two lines coincide. $\endgroup$
    – Blue
    Dec 27, 2013 at 7:35
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Hint:

  • You want to prove that $H$ lies on the bisector of $AC$.
  • $A$ is the orthocenter of $\triangle BDH$.
  • Prove the following lemma:

    For any triangle $XYZ$, the three reflections $H_{XY}$, $H_{YZ}$, $H_{ZX}$ of an orthocenter $H$ (with respect to sides) are concyclic with $X$, $Y$ and $Z$.

  • Try to reverse the following argument: $|\angle ABH| = |\angle CBH|$ because $|\angle ABH| = |\angle ADH|$ (because $\angle BHD$ is common to appropriate right triangles) and $|\angle CBH| = |\angle CDH|$ (assuming $BDCH$ is cyclic, depending on the position of $H$, some angles might be "flipped").
I hope this helps $\ddot\smile$

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