Interior angle sum property of a quadrilateral — lesson. Mathematics CBSE, Class 8.

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Now recall the angle sum property of a triangle that we studied earlier.

The sum of the measure of three angles of a triangle is

$180°$ .

Proof:

Consider a triangle

$ABC$ with interior angles measures

$∠1$ ,

$∠2$ and

$∠3$ .

Draw a line

$DE$ 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,

$∠5$ and

$∠CAD$ are linear pairs (Pair of adjacent supplementary angles).

$∠5 + ∠CAD = 180°$ .

That is,

$∠5 + ∠1 + ∠4 = 180°$ .

Equivalently,

$∠1 + ∠2 + ∠3 = 180°$ .

It states that the total measures of the three angles of a triangle is

$180°$ .

Let's extend the angle sum property to the quadrilateral.

The sum of all the angles of a quadrilateral is

$360°$ .

Proof:

Let

$ABCD$ be a quadrilateral.

Cut the quadrilateral into two triangles by drawing one of its diagonals

$AC$ .

It is obvious from the figure that

$∠1+∠2 =∠A$ and

$∠3+∠4 =∠C$ .

We know that the sum of all the angles of a triangle is

$180°$ .

Consider the triangle

$ABC$ .

$∠1 + ∠4 + ∠B =180°$ .

and

$∠2 + ∠3 + ∠D =180°$ .

Adding the above equations will result as follows:

$∠1 + ∠4$

$+ ∠B$

$+ ∠2$

$+ ∠3$

$+ ∠D$

$=$

$180° + 180°$ .

Arranging the angles in order becomes:

$(∠1 + ∠2)$

$+ ∠B +$

$(∠3 + ∠4)$

$+ ∠D$

$= 360°$ .

Substitute the known values

$∠1+∠2 =∠A$ and

$∠3+∠4 =∠C$ ,

$∠A$

$+ ∠B$

$+ ∠C$

$+ ∠D$

$= 360°$ .

Thus, the sum of the angles of a quadrilateral is

$360°$ .

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1. Interior angle sum property of a quadrilateral

Theory: