**How Do You Find The Angle Of An Isosceles Triangle**

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

**Given:** In ∆ABC, AB = AC

**To Prove:** ∠B = ∠C

**Construction:** Draw AD, bisector of ∠A

∴ ∠1 = ∠2

**Proof:** In ∆ADB & ∆ADC

AD = AD (Common)

∠1 = ∠2 (by construction)

AB = AC

By SAS, ∆ADB ≅ ∆ADC

∴ ∠B = ∠C (c.p.c.t.) Proved.

**Note:** Other result: ∠ADB = ∠ADC (c.p.c.t.)

But ∠ADB + ∠ADC = 180° (linear pair)

∴ ∠ADB = ∠ADC = 90° ⇒ AD ⊥ BC

and BD = DC (c.p.c.t.) ⇒ AD is median

∴ we can say AD is perpendicular bisector of BC or we can say in isosceles ∆, median is angle bisector and perpendicular to base also.

**Read More:**

- Angle Sum Property of a Triangle
- Median and Altitude of a Triangle
- Areas of Two Similar Triangles
- Area of A Triangle
- To Prove Triangles Are Congruent
- Criteria For Similarity of Triangles
- Construction of an Equilateral Triangle
- Classification of Triangles

**Angle Of An Isosceles Triangle Example Problems With Solutions**

**Example 1: **Find ∠BAC of an isosceles triangle in which AB = AC and ∠B = 1/3 of right angle.

**Solution:**

**Example 2: **In isosceles triangle DEF, DE = EF and ∠E = 70° then find other two angles.

**Solution:**

**Example 3: **∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see fig.). If AD is extended to intersect BC at P. Show that

(i) ∆ABD ≅ ∆ACD

(ii) ∆ABP ≅ ∆ACP

(iii) AP bisects ∠A as well as ∠D

(iv) AP is the perpendicular bisector of BC.

**Solution:**

**Example 4: **Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆PQR (see figure ). Show that:

(i) ∆ABM ≅ ∆PQN (ii) ∆ABC ≅ ∆PQR

**Solution:**