UPSKILL MATH PLUS
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Learn moreAngle sum property of a triangle:
The sum of the three angles of any triangle is \(180°\).
Example:
Consider a triangle \(ABC\) with interior angles measures \(∠1\), \(∠2\) and \(∠3\). Draw a line \(DE\) through vertex \(A\) parallel to \(BC\).
Now the angle formed by the parallel line \(DE\) with the triangle \(ABC\) is \(∠4\) and \(∠5\).
Since \(DE\) is parallel to \(BC\), using the alternate interior angle property \(∠2\) must equal to \(∠4\).
Similarly, \(∠3\) must be equal to \(∠5\).
That is \(∠2 = ∠4\) and \(∠3 =∠5\).
As \(DE\) is a straight line, \(∠5\), \(∠4\), and \(∠1\) are linear pairs.
That is, \(∠5 + ∠1 + ∠4 = 180°\)
Equivalently, \(∠1 + ∠2 + ∠3 = 180°\).
It states that the total measure of the three angles of a triangle is \(180°\).
Exterior angle of a triangle:
The angle formed between the extension of a side of a triangle and the other side is called an exterior angle of the triangle.
In this figure, for the vertex \(C\), the interior angle is \(∠ACB = c\), and the exterior angle is \(∠ACD = d\).
Forms of exterior angles in a triangle:
Exterior angles can be formed for a triangle in many ways. They are as follows:
Exterior angle property:
Statement:
An exterior angle of a triangle is equal to the sum of its opposite interior angles.
By the property, we can \(\angle ACD = \angle ABC + \angle BAC\)
Proof:
Given:
Consider a triangle \(ABC\) with extended line \(CD\) forming an exterior angle at vertex \(C\).
To prove:
\(∠ACD\) \(=\) \(∠BAC\) \(+\) \(∠ABC\)
Proof:
We will prove this using the angle sum property of the triangle.
The angle sum property states that,
The sum of the measures of the three angles of a triangle is \(180°\).
By the property, \(\angle BAC + \angle ABC + \angle ACB = 180°\) ..........(1)
Consider the line \(BD\).
Here, \(\angle ACB\) and \(\angle ACD\) are linear angles.
We know that,
The sum of all angles on a straight line is \(180°\).
\(\Rightarrow\) \(\angle ACB + \angle ACD = 180°\) ..........(2)
Equating (1) and (2), we have:
\(\angle BAC + \angle ABC + \angle ACB = \angle ACB + \angle ACD\)
\(\angle BAC + \angle ABC + \angle ACB - \angle ACB = \angle ACD\)
\(∠BAC\) \(+\) \(∠ABC\) \(=\) \(\angle ACD\)
Hence, the proof.
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