**MEDIAN**

A median of a triangle is a segment from a vertex to the midpoint of the opposite side. The 3 medians of triangle ABC are shown above. The median divides the triangles into 2 equal parts since the 2 triangles share the same altitude and base.

When we construct 2 medians of a triangle together, they interesect at a point G. The third median passes through the same point G, called the centroid, and traingle ABC is now equally divided into 6 parts.

**Proof that triangle ABC is now equally divided into 6 parts:**

The following pairs of triangles share the same base and altitudes hence they have the same area:

Triangle 1 and 2

Traingle 3 & 4

Triangle 5 and 6

Triangle ABD and ACD

Triangle CED and CEA

Triangle BAF and BCF

Sum of regions (1 + 2 + 5 ) = Sum of regions (3 + 4 + 6) and area of region 5 = area of region 6,

implies

Sum of area of regions (1 + 2 ) = Sum of area of regions (3 + 4)

That is,

2 Area of region 1 = 2 Area of region 3 , hence, area of region1 = area of region 3

2 Area of region 2 = 2 Area of region 4, hence, area of region 2 = area of region 4

2 Area of region 1 = 2 Area of region 4, hence, area of region 1 = area of region 4

2 Area of region 2 = 2 Area of region 3, hence, area of region 2 = area of region 3

Sum of regions (2 + 5 + 6 ) = Sum of regions (1 + 3 + 4)

Since, area of region2 = area of region 1 and area of region3 = area of region 4,

Area of region 5 = Area of region 3

Area of region 5 = Area of region 4

Area of region 6= Area of region 3

Area of region 6= Area of region 4

Sum of regions (4 + 5 + 6 ) = Sum of regions (1 + 2 + 3)

using the same reasoning as above, we have area of region 5 = area of region 2

Hence

Area of region 1 = 3 = 4 = 6 = 5 = 2 = 1.

**ORTHOCENTER**

When we change the dimensions of the traingle, the centroid always remain in the triangle. Next, we construct the orthocenter (H) of a triangle. The orthocenter of a triangle is the common intersection of the 3 lines containing the altitudes. An altitude is a perpendicular segment from a vertex to the line of the opposite side and H lies on the lines extended along the altitudes. When we shift or change the position of A, B or C respectively, G remains in the triangle but H does not.

I investigated the movment of H. I constructed a line DE parallel to BC and trace the movement of H when A moves along DE. H seems to form a parabola with directrix DE. Next, I attempted to locate the exact position of H.

H is the intersection of line k and l. The equation for line AC is easy to find using the points A(b,c) and C(a,0).

Gradient of AC is

Gradient of line l is

Equation of line l

Equation of line k is given by x =b

Hence, H takes up the following values:

It simply means, for any given triangle, the x-coordinate of H follows the x-coordinate of one of the vertex of the triangle. The general expression for the y-coordinate of H suggests that y takes any inside and outside the triangle. The equation of the parabola is

Ckeck:

When x=a,

To locate the vertex:

The vertex is at

and

**CIRCUMCENTER (C)**

The circumcenter C is a point where the perpendicular bisectors of each side of the triangle meet. from our construction, we note that C may be outside the triangle.

We want to show that the 3 bisectors of the sides are concurrent.

Equation of segment AB

Midpoint of AB is

The perpendicular bisector of AB has a gradient of and it passes through the midpoint of AB. Hence,

where d is the y-intercept. Then,

We do the same to find the equation for the perpendicular bisector of AD. We get

Similarily, the perpendicular bisector of BD is

The 3 lines intercept at

The 3 bisectors are thus concurrent.

**Incenter**

The three angle bisectors of the internal angles of a triangle are concurrent.

**Nine-Point Circle**

We want to prove that H, G and C are collinear and HG = 2GC.

Coordinates of C:

Coordinates of H:

Coordinates of G:

Consider 2 points C and H:

Gradient of segment CH is

y-intercept of segment CH is given by

Equation of CH is

It can be shown easily that Point G lies on the above line. Hence, HGC are collinear.

**Medial Traingle**

If we take any triangle, construct a triangle connecting the 3 midpoints of the sides, we obtain the medial triangle. It is similiar to the original triangle and one quarter its area.

**Proof**

Area of Triangle ABC is

Area of Triangle DEF is

This can be proved using paper folding exercise.

**Interior Triangle**

The shaded triangle is

(a) similar to triangle ABC

(b) congruent to the medial triangle