Show that in an isosceles triangle , angles opposite to equal sides are equal
Answers
Step-by-step explanation:
Isosceles Triangle Theorems and Proofs
Theorem 1: Angles opposite to the equal sides of an isosceles triangle are also equal.
Proof: Consider an isosceles triangle ABC where AC = BC.
We need to prove that the angles opposite to the sides AC and BC are equal, that is, ∠CAB = ∠CBA.
Isosceles Triangle
We first draw a bisector of ∠ACB and name it as CD.
Now in ∆ACD and ∆BCD we have,
AC = BC (Given)
∠ACD = ∠BCD (By construction)
CD = CD (Common to both)
Thus, ∆ACD ≅∆BCD (By SAS congruence criterion)
So, ∠CAB = ∠CBA (By CPCT)
Hence proved.
Theorem 2: Sides opposite to the equal angles of a triangle are equal.
Proof: In a triangle ABC, base angles are equal and we need to prove that AC = BC or ∆ABC is an isosceles triangle.
Isosceles Triangle Theorem 2
Construct a bisector CD which meets the side AB at right angles.
Now in ∆ACD and ∆BCD we have,
∠ACD = ∠BCD (By construction)
CD = CD (Common side)
∠ADC = ∠BDC = 90° (By construction)
Thus, ∆ACD ≅ ∆BCD (By ASA congruence criterion)
So, AC = BC (By CPCT)
Or ∆ABC is isosceles.