The addition of all three interior angles of a triangle is $180^\circ$.

Three interior angles are formed in any type of triangle and the addition of them is always $180^\circ$ geometrically.

For example, $\alpha$, $\beta$ and $\gamma$ are three interior angles of the triangle.

$\alpha + \beta + \gamma = 180^\circ$

$\Delta RST$ is a triangle and its sides are $\overline{RS}$, $\overline{ST}$ and $\overline{RT}$. The intersection of three sides form three interior angles and they are denoted by $\alpha$, $\beta$ and $\gamma$. Extend the side $\overline{RT}$ externally and assume it passes through point $U$. Draw a line from point $T$ and it should be parallel to the side $\overline{RS}$. Assume, it passes through the point $V$.

The sides $\overline{RS}$ and $\overrightarrow{TV}$ are parallel lines and they both intersected by the ray $\overrightarrow{RU}$. In other words, the ray $\overrightarrow{RU}$ becomes a transversal to the parallel lines $\overline{RS}$ and $\overrightarrow{TV}$.

Therefore, the $\angle SRT$ and $\angle VTU$ are corresponding angles but the corresponding angles are equal when a straight line intersects two parallel lines to form a transversal.

$\angle SRT = \angle VTU$

The angle $\angle SRT$ is $\beta$, So, the angle $\angle VTU$ should also be $\beta$.

$\angle SRT = \angle VTU = \beta$

The side $\overline{ST}$ is another transversal to the parallel lines $\overline{RS}$ and $\overrightarrow{TV}$.

The $\angle RST$ and $\angle STV$ are two interior alternate angles. It is proved that the interior alternative angles are equal when a straight line intersects two parallel lines to form a transversal.

$\angle RST = \angle STV$

The interior angle $\angle RST$ is $\alpha$.

$\angle RST = \angle STV = \alpha$

The angle $\angle RTV$ is a straight angle and it is $180^\circ$ exactly.

The straight angle $\angle RST$ can be obtained by adding the angles $\angle RTS$, $\angle STV$ and $\angle VTU$.

$\angle RST = 180^\circ$

$\implies \angle RTS + \angle STV + \angle VTU = 180^\circ$

$\implies \gamma + \alpha + \beta = 180^\circ$

$\therefore \,\, \alpha + \beta + \gamma = 180^\circ$

But $\alpha$, $\beta$ and $\gamma$ are interior angles of the triangle $RST$. Therefore, it is proved that summation of interior angles of the $\Delta RST$ is $180^\circ$. This property is true for all the triangles.

This property can be verified by constructing a triangle.

- Draw a straight line of $10$ centimeters by using a ruler. The endpoints of the line are called as points $O$ and $P$.
- Take compass and adjust the distance between points of needle and pencil lead to $8$ centimetres by considering the measuring system of ruler. Now draw an arc above the straight line $\overline{OP}$ from point $O$.
- Take compass and set the distance from needle point to pencil lead point to $7$ centimetres. Draw an arc above the straight line $\overline{OP}$ from point $P$.
- The arcs from point $O$ and point $P$ are intersected at point $Q$. Now join points $O$ and $Q$, and then points $P$ and $Q$ to construct a triangle, known as $\Delta OPQ$.

In $\Delta OPQ$, $O$, $P$ and $Q$ are three vertices. $\overline{OP}$, $\overline{OQ}$ and $\overline{PQ}$ are three sides of the triangle. Three interior angles are formed by the meeting of the three sides. The angles of the all three interior angles are unknown. So, measure each angle by using protractor.

$\angle QOP$ is one interior angle, which is formed by the sides $\overline{OP}$ and $\overline{OQ}$. It is measured as $44^\circ$ exactly.

$\therefore \,\, \angle QOP = 44^\circ$.

$\angle OPQ$ is another interior angle, which is formed by the meeting of sides $\overline{OP}$ and $\overline{PQ}$. Measure it and it is measured as $52.5^\circ$ exactly.

$\therefore \,\, \angle OPQ = 52.5^\circ$

$\angle PQO$ is another interior angle of the triangle $OPQ$. It is formed by the meeting of the sides $\overline{OQ}$ and $\overline{PQ}$. The angle is measured as $83.5^\circ$ exactly.

$\therefore \,\, \angle PQO = 83.5^\circ$

Now, add all three interior angles of the triangle $OPQ$.

$\angle QOP + \angle OPQ + \angle PQO = 44^\circ + 52.5^\circ + 83.5^\circ$

$\therefore \,\, \angle QOP + \angle OPQ + \angle PQO = 180^\circ$

It is proved that the summation of the all three interior angles is $180^\circ$ exactly in any type of triangle.

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