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We Can Show That The Orthocentre

The orthocentre, centroid, and circumcentre of any triangle are always collinear. This is proven through 10 steps: (1) constructing lines through the centroid and orthocentre, (2) showing they intersect at the circumcentre, (3) using properties of the centroid and circumcentre to show the orthocentre must be where the lines intersect. Therefore, the three points always lie on the Euler line. Additionally, the nine-point centre lies on the Euler line midpoint between the orthocentre and circumcentre.

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305 views3 pages

We Can Show That The Orthocentre

The orthocentre, centroid, and circumcentre of any triangle are always collinear. This is proven through 10 steps: (1) constructing lines through the centroid and orthocentre, (2) showing they intersect at the circumcentre, (3) using properties of the centroid and circumcentre to show the orthocentre must be where the lines intersect. Therefore, the three points always lie on the Euler line. Additionally, the nine-point centre lies on the Euler line midpoint between the orthocentre and circumcentre.

Uploaded by

sagar
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© © All Rights Reserved
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We can show that the orthocentre, circumcentre and the centroid of any triangle are always

collinear in the following way:-

Let the centroid be (G),

the orthocenter (H) and

the circumcenter (C).

Step1:- Let X be the


midpoint of EF.
Construct the
median DX. Since G is
the centroid, G is
on DX by the definition of centroid. Also, construct the altitude DM. Since H is the orthocenter, H is
on DM by the definition of orthocenter. Therefore, DM meets EF at a right angle.

Step2:- Construct a line through points C and G so that it intersects DM. Note, it appears that
line CG intersects DM at the point H. However, this has not been proven yet. So, label the point of
intersection H'.

Step3:- Construct CX. Since C is the


circumcenter, CX meets EF at a right angle by
the definition of circumcenter.

Step4:- Since DM and CX both meet EF at


right angles, DM is parallel to CX. Then, the
median DX is a transversal that
cuts DM and CX.

Step5:-

Step6:- Since G is the centroid, DG = (2/3)DX and GX = (1/3)DX. Therefore, DG = 2GX.

Step7:-
Step8:- Thus, H' is located at the intersection of DM and GC and is in fact on the line GC such
that GH' = 2GC. Thus, G, H', and C are collinear. Similarly, H' can be proved to be located on the
altitudes constructed from vertices E and F so that GH' = 2GC. Therefore, H' lies on all three
altitudes.

Step9:- Thus, H' is the orthocenter because it is lies on all three altitudes. Yet, by the given
hypothesis, H is the orthocenter. Thus, H' = H.

Step10:- Therefore, G, H, and C are


collinear, and GH = 2GC.

The line that connects the centroid (G), the


orthocenter (H), and the circumcenter (C) is
called the Euler Line.

In case of an equilateral triangle,

The Nine-point center is the center of the


nine-point circle. The Nine-Point Circle of
triangle ABC with orthocenter H is the circle
that passes through the feet of the altitudes
HA, HB and HC to the three sides, the midpoints
MA, MB and MC of those sides, and the Euler
Points EA, EB and EC, which are the midpoints of
the segments AH, BH, and CH, respectively. Euler line is the line passing through the orthocenter H, the
nine-point center N, the centroid G, and the circumcenter O of any triangle ABC.
The nine-point center N is the midpoint of the line HO.
The distance from the orthocenter H to the centroid G is twice the distance from the
circumcenter O to the centroid G.
The nine-point center N is the circumcenter of the medial triangle M AMBMC.
The nine-point center N is the circumcenter of the orthic triangle H AHBHC.
The nine-point circle is also known as Euler's circle and Feuerbach's circle. Leonhard Euler showed in
1765 that the nine-point circle bisects any line from the orthocenter to a point on the circumcircle. In
1822 Karl Feuerbach discovered that any triangle's nine-point circle is externally tangent to that
triangle's three excircles and internally tangent to its incircle.

Mirror image of orthocenter about any side of triangle lies on its circumcircle.

Ceva's theorem

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