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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.
Solution
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.
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