Question

draw an equilateral triangle of sides 5 cm find its centroid orthocentre and circumcentre . what conclusion can you make ?​

Answer 1

Answer:

it will come At the centre of equilatrel triangle

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hope this helps you

Answer 2

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:-

Therefore, ∠GDH’ = ∠GXC, since they are alternate interior angles. Also, ∠DGH’ = ∠XGC, because they are vertically opposite angles.

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

Step7:-

Since ∠GDH’ = ∠GXC and ∠DGH’ = ∠XGC, ΔDGH’ ~ ΔXGC, by A-A similarity criterion. Therefore, GH’ = 2GC; since DG = 2GX.

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 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,

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