Question

In the adjoining figure, ABC is a triangle in which AB = AC. If D and E are points on AB and AC respectively such that AD = AE, show that the points B, C, E and D are concyclic.  

Answer 1

Given:
AD = AE …(1)
AB = AC …(2)
Subtracting AD from both sides, we get:
⟹ AB – AD = AC – AD
⟹ AB – AD = AC - AE (Since, AD = AE)
⟹ BD = EC …(3)
Dividing equation (1) by equation (3), we get: 

`(AD)/(DB)=(AE)/(EC)` 

Applying the converse of Thales’ theorem, DE‖BC
⟹ ∠DEC + ∠ECB = 180° (Sum of interior angles on the same side of a
                                                      Transversal Line is 0°) 

⟹ ∠DEC + ∠CBD = 180° (Since, AB = AC ⟹ ∠B = ∠C)
Hence, quadrilateral BCED is cyclic.
Therefore, B,C,E and D are concylic points.