Question

10. In any triangle ABC, if the angle bisector of ZA and perpendicular bisector of BC
intersect, prove that they intersect on the circumcircle of the triangle ABC.​

Answer 1

$\huge\blue\bigstar\pink{❥ᴀ᭄ɴswer★}$

∠BAE=∠CAE {Angles inscribed in the same arc}

∴arc BE=arc CE

∴arc BE=arc CE∴ chord BE≅ chord CE ___(1)

(1)In △BDE and △CDE,

(1)In △BDE and △CDE, BE=CE {from (1)}

BD=CD {ED is perpendicular bisector of BC}

DE=DE {common side}

∴△BDE≅△CDE {SSS test of congruence}

⇒∠BDE=∠CDE {C.P.C.T}

Also, ∠BDE+∠CDE=180∘ {angles in linear pair}

∴ ∠BDE = ∠CDE = 90∘

90∘

90∘ Therefore, DE is the right bisector of BC.

Answer 2

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