Isosceles triangle theorem is one of the most important theorems in geometry. Greeks and Egyptians have used this theorem since ancient times, and some real-life examples are beautiful architecture and pyramids. Let’s dive into some of the essential concepts of the theorem.
What is a theorem?
We must know what does a theorem means before jumping into the isosceles triangle theorem. A theorem is a statement that needs to be proved to demonstrate the relation between concepts. In short, a theorem is a set of rules or principles that needs to be proved to determine whether the relationship between concepts is true or false.
Let’s try to understand it with an example:
If we have to prove that the square of the hypotenuse side (in this case AC) is equal to the square of the other two sides of the triangle (i.e., Side AB and Side BC). So to prove this statement, we have to write down some steps or rules, which is what we call a theorem.
What is an Isosceles triangle?
An isosceles triangle is a triangle in which two sides of the triangle have equal length.
According to the above figure in △ BAC, side AB, and side BC are equal in length, denoted by two black lines. As we see, l(AB)= 2 and l(AC)= 2; therefore, the △ BAC is an isosceles triangle.
Properties of an isosceles triangle
We now understand the definition of an isosceles triangle. Let’s have a look at some of its properties.
|1. The equal sides of an isosceles are known as legs.|
|2. The third unequal side of an isosceles triangle is known as the base.|
|3. The angle between the equal side of the triangle is known as the vertex angle.|
|4. The angle made by the base (i.e., the unequal side) is called base angles.|
|5. The angles which are opposite to the equal sides of the triangle are always equal.|
|6. All the three angles inside the isosceles triangle are acute angles.|
|7. The sum of all the angles in an isosceles triangle is always 180°|
Isosceles triangle theorem
Statement: If two sides of a triangle are congruent, then the angles opposite them are congruent.
(Note: the symbol ‘≅‘ means congruent)
Given: In △ ABC, side AB ≅ side AC
To prove: ∠ABC ≅ ∠ACB
Construction: Draw the bisector of ∠BAC, which intersects side BC at point D.
|Proof:||In △ ABD and △ ACD|
|Seg AB ≅ Seg AC||Given|
|∠BAD ≅ ∠CAD||According to construction|
|Seg AD ≅ Seg AD||Common side|
|∴ ∆ ABD ≅ ∆ACD|
|∴ ∠ ABD ≅ ∠ ACD|
|∴ ∠ ABC ≅ ∠ ACB|
Hence proved that if two sides of a triangle are congruent (Seg AB ≅ Seg AC), then the angles opposite to them are also congruent (∠ABC ≅ ∠ACB).
The converse of isosceles triangle theorem
Theorem: If two angles of a triangle are congruent, then the sides opposite to them are congruent.
Given: In ∆ QPR, ∠ QPR ≅ ∠ PRQ
To prove: Side PQ ≅ Side PR.
Construction: Draw the bisector of ∠ P intersecting side QR at point M.
|Proof:||In ∆ PQM and ∆ PRM|
|∠ PQM ≅ ∠ PRM||Given|
|∠ QPM ≅ ∠ RPM|
|Seg PM ≅ Seg PM||Common side|
|∴ ∆ PQM ≅ ∆ PRM|
|∴ Seg PQ ≅ Seg PR|
Hence proved that if two angles of a triangle (∠ PQM ≅ ∠ PRM) are congruent, then the sides opposite to them are congruent (Seg PQ ≅ Seg PR).
1. In the given isosceles ∆ PQR, find the measurement of side QM and the area of ∆ PMQ.
In an isosceles triangle, the perpendicular from the vertex angle bisects the base.
So according to the given figure:
|QM = MR|
|∴ QM = 3 units.|
In ∆ PMQ, the base is 3 units and the height is 4 units.
|∴ Area of ∆ PMQ||= ½ * b * h|
|= ½ * 3 * 4|
|= 6 units|
2. ∆ CAB is an isosceles triangle with ∠ A being the vertex angle. Find the value of n and then classify ∆ CAB.
Given: ∠ A is the vertex angle of the isosceles triangle.
To find: The value of n and classify ∆ CAB.
We know that if ∠ A is the vertex angle, then ∠ B and ∠ C are the base angles according to the theorem of an isosceles triangle.
Therefore, ∠ B and angle ∠ C are congruent to each other.
|∴ Seg CA = Seg BA||By theorem of the isosceles triangle|
|∴ 3n + 46 = 12n + 10|
|∴ 46 – 10 = 12n – 3n|
|∴ 36 = 9n|
|∴ n = 4|
Value of n = 4.
Now that we know the value of n, let’s evaluate the value of all the segments of ∆ CAB.
|∴ Seg CA||= 3n + 46|
|∴ Seg CA||= 3(4) + 46|
|∴ Seg CA||= 12 + 46|
|∴ Seg CA||= 58|
|∴ Seg CB||= 8n + 21|
|∴ Seg CB||= 8(4) + 21|
|∴ Seg CB||= 32 + 21|
|∴ Seg CB||= 53|
|∴ Seg AB||= 12n + 10|
|∴ Seg AB||= 12(4) + 10|
|∴ Seg AB||= 48 + 10|
|∴ Seg AB||= 58|
Therefore as CA = AB = 58, we have confirmed that ∆ CAB is an isosceles triangle.
Unsolved examples for practice:
- In the given ∆ BAC, ∠A is the vertex angle. Find the value of x and evaluate the values of all the segments of ∆ BAC.
2. In the given ∆ QPM ∠P=(5a+9)° is the vertex angle. The value of ∠ M = 28°. Find the value of a and the measurement of ∠ P.
If you have any doubts regarding the article or the examples, please post them in the comments section.