Properties of Triangles:

a. Angle Sum Property

Theorem: The sum of the angles of a triangle is 180°.

Proof:

Given: $\Delta ABC$

To Prove: $\angle ABC+\angle BCA+\angle CAB=180^{\circ}$

First draw a line $DE$ parallel to $BC (DE \parallel BC)$.

$AB \ and\ AC$ are the transversals.

Since $DE \parallel BC$,

$\angle ABC=\angle BAD$ (Alternate Angles)……………………….i)

Similarly,

$\angle BCA=\angle CAE$ (Alternate Angles)…………………………ii)

Adding i) and ii), we get

$\angle ABC+\angle BCA=\angle BAD+\angle CAE$ …………………………iii)

Adding $\angle BAC$ on both sides of iii) we get the following

$\angle ABC+\angle BCA+\angle BAC=\angle BAD+\angle CAE+\angle BAC$

Since $\angle BAD+\angle CAE+\angle BAC=180^{\circ}$  (Angles on a straight line)

Hence

$\angle ABC+\angle BCA+\angle BAC=180^{\circ}$

or $\angle A+\angle B+\angle C=180^{\circ}$

b. Exterior Angle Property

Theorem: If one side of the triangle is produced, then the exterior angle so formed is equal to the sum of the interior opposite angles.

Proof:

To Prove: $\angle ACD=\angle CAB+\angle BAC$

Given: Given: $\Delta ABC$

Extend $BC\ to\ D$. Also draw a like $CE \parallel AB.\ AC\ \&\ BC$ are the transversals.

Therefore, $\angle ACE=\angle BAC$ (Alternate Angles)

Similarly,$\angle DCE=\angle ABC$ (Corresponding Angles)

$\angle ACD=\angle ACE+\angle ECD=\angle ABC+\angle BAC$

or $\angle ACD=\angle ABC+\angle BAC$

Hence proved, that if one side of the triangle is produced, then the exterior angle so formed is equal to the sum of the interior opposite angles.