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Triangle ABC is inscribed in a circle $C(O, r )$. The angle bisectors of A, B, and C intersect the circle in D, E, and F respectively. Show that the angle bisectors of angles A, B, C of $triangle ABC$ are the altitudes of triangle $DEF$.

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    $\begingroup$ Do you understand why the $\alpha,\beta,\gamma$ in $AGE$ are the same angles as the angles labelled so? Do you see why $2\alpha+2\beta+2\gamma=180^\circ$? $\endgroup$ May 15, 2018 at 20:14
  • $\begingroup$ No I'm not to sure why sorry. $\endgroup$
    – Jdinh98
    May 15, 2018 at 20:22

6 Answers 6

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Using the Angle of Intersecting Chords Theorem

Using the Inscribed Angle Theorem, $\overset{\large\frown}{AF}=2\gamma$ and $\overset{\large\frown}{AE}=2\beta$; therefore, $\overset{\large\frown}{EF}=2\beta+2\gamma$. Furthermore, $\overset{\large\frown}{DB}=2\alpha$. Since they are the angles of $\triangle ABC$, $2\alpha+2\beta+2\gamma=\pi$. Therefore, using the Angle of Intersecting Chords Theorem, we get that the angle between $\overline{FD}$ and $\overline{EB}$ is $$ \begin{align} \frac12\left(\overset{\large\frown}{EF}+\overset{\large\frown}{DB}\right) &=\frac12\left(2\beta+2\gamma+2\alpha\right)\\[6pt] &=\frac\pi2 \end{align} $$ The other intersections are similar.


Proof of the Angle of Intersecting Chords Theorem

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Using the Inscribed Angle Theorem, $\overset{\large\frown}{AB}=2\angle ACB$ and $\overset{\large\frown}{CD}=2\angle CAD$. The Exterior Angle Theorem then says that $\angle AOB=\angle ACB+\angle CAD$.

Therefore, we get the Angle of Intersecting Chords Theorem: $$ 2\angle AOB=\overset{\large\frown}{AB}+\overset{\large\frown}{CD} $$

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All triangles drawn from a chord will have the same angle where they touch the circumference.

For instance triangles $\triangle AEF$ and $\triangle ACF$ make the angle $\gamma$ at points $E$ and $C$, respectively. For the same reason $\angle EAC = \angle EBC = \beta$.

In $\triangle ABC$, the sum of all angles must be $180^{\circ}$. Thus $2(\alpha+\beta+\gamma)=180^{\circ}$ implying $\alpha+\beta+\gamma=90^{\circ}$. Now if you look at the right triangle $\triangle EGA$, $\angle EGA=180^{\circ}-(\alpha+\beta+\gamma)=90^{\circ}$. Thus, $DG \perp EF$.

Similarly, if the intersection points of lines $ED$ and $CF$ are named $M$ and that of $EB$ and $DF$ are named $N$, the result can be proved for the other two bisecting lines by considering the traingles $\triangle EMC$ and $\triangle BDN$.

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In $\triangle DGF$, angles subtended by a common chord in a circle are equal, you will get $\angle DFC=\alpha$ and $\angle CFE=\beta$ and summing them $\angle F=\alpha +\beta$.
Now $\angle ADF=\gamma$ for same reason.
$\Rightarrow$ $180^\circ - (\angle GDF+\angle DFG)=\angle DGF$
$\Rightarrow$ $180^\circ -(\alpha +\beta +\gamma)=\angle DGF=90^\circ$
Similarly all angle bisectors will be perpendiculars for $\triangle DEF$

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Let $I$ be the incenter of $ABC$. It is well know'n that $D$ is a center of circle through $BCI$. The same is valid for $E$ and circle $ACI$. So $CI$ is radical line for circles $BCI$ and $ACI$ so $CI$ is perpendicular to line between their centers i.e. $DE$.

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$$ 2 \alpha + 2 \beta + 2 \gamma = 2 \pi \rightarrow \alpha + \beta + \gamma = \pi $$ Consider $\Delta EAG \, $

$$\angle HGA + \alpha + \beta + \gamma = 2 \pi \rightarrow\, \angle HGA = \pi$$

$$ \angle \, bisector \, AGD \, is \perp^{r} to \, line \,EGF... $$

So also for altitudes $ EF,BE$ making unlabeled point of congruence incenter of $\Delta ABC$ the same as orthocenter of $ \Delta DEF$.

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(Name the intersection of AB and EF as I)

Join AF

Angle EFA = Beta

Angle BAF = Gamma

By exterior angle property,

Angle GHA = Angle GIA = Beta + Gamma

Angle HAG = Angle IAG = Alpha

From these we can say that Angle AGH = Angle AGI

Also Angle AGH + Angle AGI = 180

Thus, AD is altitude of ∆DEF

Similarly you can show others

(Sorry for not writing symbols)

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