The birth of a True Equilateral Triangle from Isosceles parents is the subject of rejoicing in our country from many furlongs around.
Edwin Abbott
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There exist several types of triangles, each characterized by three angles. While the specific angles may vary, there is a consistent relationship among them. This is encapsulated in the Angle Sum Property, which asserts that the sum of the three angles in any triangle is constant.
The total measure of the three angles of a triangle is 180°.
In this blog post, we will explore several methods to substantiate the above statement. While the ways provided is not exhaustive, we encourage you to share any alternative approaches you may have discovered by reaching out to us via email. Your contributions and perspectives are valuable to us.
To validate the given statement, we can employ the approach of measuring all three angles in various triangles using a protractor. Considering the six different types of triangles, we proceed to draw each one. Subsequently, we measure the angles of each triangle and calculate their sum. It is evident that the total sum consistently equals 180°.
We know that the sum of the interior angles of a regular polygon with n sides = 180(n ‒ 2)°. In the case of a triangle, n = 3. Substituting the value of n, we find that
the sum of the interior angles of a triangle = 180(3 ‒ 2)° = 180°.
Draw a triangle and label its angles as 1, 2, and 3. Cut along the three angles. Rearrange them as shown in the following figure. The three angles now form a single angle. This angle constitutes a straight angle and, therefore, has a measure of 180°. Thus, the sum of the measures of the three angles of a triangle is 180°.
The same fact can be observed in a different way also. Take three copies of any triangle, say ΔABC. Arrange them as shown in the following figure. What is your observation regarding do you observe about ∠1 + ∠2 + ∠3? Can you see the exterior angle property?
The sum of the exterior angles of a triangle is 360°. If we call them as ∠1, ∠2, and ∠3, then
∠1 + ∠2 + ∠3 = 360°.
It is clear that
∠1 + ∠CAB = 180°;
∠2 + ∠ABC = 180°;
∠3 + ∠BCA = 180°.
Adding the respective sides of the above relations, we find that
∠1 + ∠CAB + ∠2 + ∠ABC + ∠3 + ∠BCA = 540°
Or, ∠1 + ∠2 + ∠3 + ∠ABC + ∠BCA + ∠CAB = 540°
Or, ∠ABC + ∠BCA + ∠CAB = 180°
Given triangle is ABC. We need to prove that ∠A + ∠B + ∠C = 180°.
Construction. Extend the line BC to the point D and now draw a straight line CE parallel to the side AB.
Proof. Since AB ǁ CE, we have
∠BAC = ∠ACE (alternate angles);
∠ABC = ∠ECD (corresponding angles).
Since ∠ACB + ∠ACE + ∠ECD is a straight angle, on substitution, we find that ∠ACB + ∠ACE + ∠ECD = 180°.
Using exterior angle property, we can prove the angle sum property. Given triangle is ABC. We need to prove that ∠A + ∠B + ∠C = 180°.
Construction. Extend the line BC to the point D.
Proof. Exterior angle ∠ACD is equal to the sum of two opposite interior angles.
Therefore, ∠ACD = ∠ABC + ∠BAC.
Adding ∠ACB to both sides, we find that
∠ACD + ∠ACB = ∠ABC + ∠BAC + ∠ACB
Clearly, ∠ACD + ∠ACB is a straight angle, hence measures 180°.
Therefore, ∠ABC + ∠BAC + ∠ACB = 180°, as expected.
Consider a triangle ABC and mark the angles as ∠1, ∠2 and ∠3, as shown in figure i. Then make two copies of the triangle and arrange them as shown in figure ii. Continuing this way, we finally make a figure like iii, where O is the centre.
Certainly, we find a complete angle at O which is equal to 360°. Since the vertically opposite angles are equal, we find the angles as ∠1, ∠1, ∠2, ∠2, ∠3, ∠3. Of course,
∠1 + ∠1 + ∠2 + ∠2 + ∠3 + ∠3 = 360°
Or, ∠1 + ∠2 + ∠3 = 180°.
Consider the triangle ABC. Draw a line MN parallel to side AB of △ABC that goes through point C (the line is unique because of the fifth postulate). Since AB ǁ MN, it follows that
∠CAB = ∠ACM (alternate angles);
∠CBA = ∠BCN (alternate angles).
Since ∠CAB + ∠ACM + ∠ACM is a straight angle, it follows that
∠CAB + ∠ACM + ∠ACM = 180°
or, ∠A + ∠B + ∠C = 180°, as required.
Let ABC (labelled in anti-clockwise order) be the given triangle.
We denote the vertices by A(z1), B(z2), and C(z3). Then the three interior angles of the triangle ABC are given by
Therefore, the sum of the three angles of the triangle is
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