Theorem Angle opposite to equal sides of an isosceles triangle are equal (original) (raw)

Last Updated : 9 Feb, 2026

An isosceles triangle is a triangle that has two sides of equal length. Examples of isosceles triangles include the isosceles right triangle, the golden triangle, and the faces of bipyramids and certain Catalan solids.

**Theorem Statement: Angle opposite to equal sides of an isosceles triangle are equal.

**Proof:

1

Isosceles triangle ABC

Given, an Isosceles triangle ABC, where the length of side AB equals the length of side AC.

Therefore, AB = AC

Construction:

Let us draw the bisector of ∠A

Let D be the point of intersection of this bisector of ∠A and BC.

Therefore ,by construction ∠BAD = ∠CAD.

In ∆BAD and ∆DAC,

AB = AC (Given)

∠BAD = ∠CAD (By construction)

AD = AD (Common side in both triangle)

So, ∆BAD ≅ ∆CAD (By SAS rule)

So, ∠ABD = ∠ACD, since they are corresponding angles of congruent triangles.

So, ∠B = ∠C

Hence, Proved that an angle opposite to equal sides of an isosceles triangle is equal.

**Note: The converse of this theorem is also true. The sides opposite to equal angles of a triangle are also equal.

**Sample Problems Based on the Theorem

**Problem 1: E and F are respectively the mid-points of equal sides AB and AC of ∆ABC (see given figure). Show that BF = CE.

2

**Given:

Length of side AB = AC

**To show: BF = CE

In ∆ABF and ∆ACE,

AB = AC (Given)

∠A = ∠A (Common)

AF = AE (Halves of equal sides)

So, ∆ABF ≅ ∆ACE (SAS rule)

Since, If two triangles are congruent, their corresponding sides are equal.

Therefore, BF = CE ( by CPCT)

**Problem 2: Given an ∆ABC whose perimeter is 13 cm and ∠ABC = ∠ACB and length of side BC equals 3 cm. Find the **length of side AB and AC.

3

**Given:

BC = 3cm, Perimeter of ∆ABC = 13cm

∠ABC = ∠ACB

Since ∠ABC = ∠ACB , therefore by applying theorem, the sides opposite to equal angles of a triangle are also equal.

So, length of side AB = AC.

Let the side of AB be x.

Therefore, Perimeter = AB + BC + AC

13 = x + 3 + x ( Since, AB = AC )

13 = 2x + 3

13 - 3 = 2x

10/2 = x

Therefore x = 5

**So, the length of side AB and AC is 5 cm.

Practice Problems

  1. In triangle ABC, AB = AC and ∠B = 45°. Find ∠C.
  2. In triangle DEF, DE = DF and ∠D = 70°. Find ∠E and ∠F.
  3. In triangle GHI, GH = GI and ∠H = 30°. Find ∠I.
  4. In triangle JKL, JK = JL and ∠K = 55°. Find ∠L.
  5. In triangle MNO, MN = MO and ∠N = 75°. Find ∠O.
  6. In triangle PQR, PQ = PR and ∠P = 40°. Find ∠Q and ∠R.
  7. In triangle STU, ST = SU and ∠S = 60°. Find ∠T and ∠U.
  8. In triangle VWX, VW = VX and ∠V = 50°. Find ∠W and ∠X.
  9. In triangle YZA, YZ = YA and ∠Y = 35°. Find ∠Z and ∠A.
  10. In triangle BCD, BC = BD and ∠B = 80°. Find ∠C and ∠D.