Prove that “Angles opposite to equal sides of an isosceles triangle are equal.
Answers
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.
Answer: ABC, in which AB=AC.
=Construct AP bisector of angle A meeting BC at P.
=In ∆ABP and ∆ACP
=AP=AP[common]
=AB=AC[given]
=angle BAP=angle CAP[by construction]
=Therefore, ∆ABP congurent ∆ACP[S.A.S]
=This implies, angle ABP=angleACP[C.P.C.T]
hope it helps
Step-by-step explanation: