Given: AD = AE …(i)
& AB = AC …(ii)
Subtracting AD from both sides of equation (ii), we get
AB – AD = AC – AD
⇒ AB – AD = AC – AE [from equation (i)]
⇒ DB = EC [∵ AB – AD = DB & AC – AE = EC] …(iii)
Now, divide equation (i) by (iii), we get
By converse of Thale’s theorem, we can conclude by this equation that DE ∥ BC.
So, ∠DEC + ∠ECB = 180° [∵ sum of interior angles on the same transversal line is 180°]
Or ∠DEC + ∠DBC = 180° [∵ AB = AC ⇒ ∠C = ∠B]
Hence, we can write DEBC is cyclic and points D, E, B and C are concyclic.
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