Points of Concurrency in a Triangle

Medians of a Triangle are Concurrent. The Point of Concurrency is the Centroid.

A median of a triangle is the line joining the midpoint of a side and the opposite vertex.


Angle Bisectors of a Triangle are Concurrent. The Point of Concurrency is the Incenter.


Altitudes of a Triangle are Concurrent. The Point of Concurrency is the Orthocenter.

An altitude is the perpendicular line drawn from a vertex on the opposite side of the triangle.


Perpendicular Bisectors of a Triangle are Concurrent. The Point of Concurrency is the Circumcenter.



Volume of Parallelopiped and Tetrahedron

A parallelopiped is a 3-D object, each of whose faces is a parallelogram.

A rectangular parallelopiped is the one, whose faces are rectangular.

A cube is a rectangular parallelopiped, whose edges are of equal length.


A tetrahedron is a 3-D object, whose all faces are triangular. It can be shown that a parallelopiped can be decomposed into 6 tetrahedra.

Image Source : https://www.dune-project.org/

For a visual proof :  1 Parallelopipded  = 6 Tetrahedra


Let there be 2 such objects, whose coterminus edges are identical. Then, the volume of the tetrahedron is {\dfrac 1 6}th of that of the parallelopiped. The expression can also be obtained using the following theorem :


Applications of Vectors in Geometry

Vectors are such a powerful tool in mathematics and physics, that many results can be proved very easily and intuitively.



Statement : If the diagonals of a parallelogram are congruent, then it is a rectangle.

Proof :

Let ABCD be the parallelogram. Let A be the origin. So, the position vectors will be from A. Diagonals are AC and BD. We have |\vec {AC}| = |\vec {BD}|.

Parallelogram ABCD


Therefore, |\vec c| = |\vec d - \vec b|. By parallelogram law of vector addition, \vec c = \vec d + \vec b. So,

{|\vec c| = |\vec d - \vec b| = |\vec d + \vec b|}

On squaring,

{\vec d \cdot \vec d - 2 \vec d \cdot \vec b + \vec b \cdot \vec b = \vec d \cdot \vec d + 2 \vec d \cdot \vec b + \vec b \cdot \vec b}


{- 2 \vec d \cdot \vec b = 2 \vec d \cdot \vec b}

This implies \vec d \cdot \vec b = 0, or \vec b is perpendicular to \vec d. Hence, \angle A is a right angle. This being a parallelogram, will have all other angles equal to 90^o and hence it is a rectangle.


Statement : The diagonals of a kite are at right angles.

Proof :



Let ABCD be the kite, with A as the origin. Clearly BC and DC are equal sides. In terms of vectors,

 {|\vec {BC}| = |\vec {DC}| \ \ i.e. \ \ |\vec c - \vec b| = |\vec c - \vec d|}

On squaring,

{\vec c \cdot \vec c - 2 \vec c \cdot \vec b + \vec b \cdot \vec b = \vec c \cdot \vec c - 2 \vec c \cdot \vec d + \vec d \cdot \vec d}

Canceling common terms,

{- 2 \vec c \cdot \vec b = -2 \vec c \cdot \vec d}

Note that AB and AD are equal. So, b^2 = d^2. So,

{\vec c \cdot (\vec b - \vec d) = 0}

Hence, \vec {AC} \cdot \vec {DB} = 0. Hence the diagonals are perpendicular.


Statement : If in a tetrahedron, edges in each of the two pairs of opposite edges are perpendicular then the edges in the third pair are also perpendicular.


In a tetrahedron, each triangle shares an edge with the other. Considering any 2 triangular faces, we are left with only 1 edge. The pair of common edge and the uncommon edge is said to be a pair of opposite edges. Let OABC be a tetrahedron. So, (OA,BC), (AB, OC) and (OB,AC) are the pairs of opposite edges. Let any 2 of them be perpendicular.


Tetrahedron OABC

\vec {OA} \cdot \vec {BC} = 0 and \vec {OC} \cdot \vec {AB} =0. Therefore,

{\vec a \cdot (\vec c - \vec b) = 0 \ \ and \ \ \vec c \cdot {\vec b - \vec a} = 0}

Expanding the brackets and then adding the equations,

{\vec a \cdot \vec c - \vec a \cdot \vec b + \vec c \cdot \vec b - \vec c \cdot \vec a =0}


{- \vec a \cdot \vec b + \vec c \cdot \vec b = 0}

This gives \vec b \cdot (\vec c - \vec a ) = 0 i.e. \vec {OB} \cdot \vec {AC} = 0. Hence OB is perpendicular to BC and these 2 form the third pair.