Question

4. Show that in an isosceles triangle, angles opposite to equal sides are equal.​

Answer 1

Answer:

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.

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. In an isosceles triangle, angles opposite to equal sides are equal.​

Answer 2

Answer:

Here we will prove that in an isosceles triangle, the angles opposite to the equal sides are equal. Given: In the isosceles ∆XYZ, XY = XZ. Construction: Draw a line XM such that it bisects ∠YXZ and meets the side YZ at M.......