In the adjoining figure, ABC is a triangle in which AB = AC. IF D and

1.	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» Given: AD = AE …(i) AB = AC …(ii) Subtracting AD from both sides, we get: ⇒ AB – AD = AC – AD ⇒ AB – AD = AC - AE (Since, AD = AE) ⇒ BD = EC …(iii) Dividing equation (i) by equation (iii), 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.

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»

Given:

AD = AE …(i)

AB = AC …(ii)

Subtracting AD from both sides, we get:

⇒ AB – AD = AC – AD

⇒ AB – AD = AC - AE (Since, AD = AE)

⇒ BD = EC …(iii)

Dividing equation (i) by equation (iii), 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.