Orthocenter and Incenter Coincide if Triangle is Equilateral
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Theorem
Let $\triangle ABC$ be an equilateral triangle.
Let $G$ be the orthocenter of $\triangle ABC$.
Then: $G$ is the incenter of $\triangle ABC$.
Proof
Draw the circumcircle of $\triangle ABC$ through points $A, B,$ and $C$.
Given: $G$ is the orthocenter of $\triangle ABC$.
By Orthocenter, Centroid and Circumcenter Coincide iff Triangle is Equilateral:
- $G$ is also the centroid and circumcenter of $\triangle ABC$.
| \(\ds AB\) | \(=\) | \(\ds AC = CB\) | Definition of Equilateral Triangle | |||||||||||
| \(\ds AD\) | \(=\) | \(\ds DB\) | Medians of Triangle Meet at Centroid | |||||||||||
| \(\ds AD\) | \(=\) | \(\ds \dfrac 1 2 AB\) | Definition of Median | |||||||||||
| \(\ds AF\) | \(=\) | \(\ds \dfrac 1 2 AC\) | Definition of Median | |||||||||||
| \(\ds AD\) | \(=\) | \(\ds AF\) | Substitution | |||||||||||
| \(\ds AG\) | \(=\) | \(\ds AG\) | Shared | |||||||||||
| \(\ds \angle ADG\) | \(=\) | \(\ds \angle AFG\) | Given both are right angles | |||||||||||
| \(\ds \triangle ADG\) | \(\cong\) | \(\ds \triangle AFG\) | Triangle Right-Angle-Hypotenuse-Side Congruence | |||||||||||
| \(\ds AG\) | \(=\) | \(\ds BG = CG\) | Radii of the same circle | |||||||||||
| \(\ds DG\) | \(=\) | \(\ds DG\) | Shared | |||||||||||
| \(\ds \triangle ADG\) | \(\cong\) | \(\ds \triangle BDG\) | Triangle Side-Side-Side Congruence | |||||||||||
| \(\ds \triangle AFG\) | \(\cong\) | \(\ds \triangle BDG\) | Common Notion 1 |
By mutatis mutandis:
all the smaller triangles are congruent.
So
- $DG = EG = FG$
By the definition of incircle:
By definition of center:
$\blacksquare$