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20/07/2021 5:01 pm
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In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
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20/07/2021 5:03 pm
Consider this diagram
Here, join BE and CE.
Now, since AE is the bisector of ∠BAC
∠BAE = ∠CAE
∴ arc BE = arc EC
This implies, chord BE = chord EC
Now, consider triangles ΔBDE and ΔCDE,
DE = DE (It is the common side)
BD = CD (It is given in the question)
BE = CE (Already proved)
So, by SSS congruency, ΔBDE ΔCDE.
∴ ∠BDE = ∠CDE
We know, ∠BDE = ∠CDE = 180°
∠BDE = ∠CDE = 90°
∴ DE ⊥ BC (hence proved).