Prove that the quadrilateral formed (if possible) by the internal angle bisectors of any
quadrilateral is cyclic
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Prove that the quadrilateral formed (if possible) by the internal angle bisectors of any
quadrilateral is cyclic
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Given a quadrilateral ABCD with internal angle bisectors AF, BH, CH
and DF of angles A, B, C and D respectively and the points E, F, G
and H form a quadrilateral EFGH.
To prove that EFGH is a cyclic quadrilateral.
∠HEF = ∠AEB ⠀⠀⠀ [Since,Vertically opposite angles] -------- (1)
Consider triangle AEB,
∠AEB + 1/2 ∠A + 1/2 ∠B = 180°
∠AEB = 180° – 1/2 (∠A + ∠ B) -------- (2)
From (1) and (2),
∠HEF = 180° – 1/2 (∠A + ∠ B) --------- (3)
Similarly, ∠HGF = 180°– 1/2(∠C + ∠ D) ------- (4)
From 3 and 4,
∠HEF +∠HGF = 360°– 1/2∠A +∠B +∠C +∠ D)
= 360° – 1/2 (360°)
= 360° – 180°
= 180°
So, EFGH is a cyclic quadrilateral since the sum of the opposite
angles of the quadrilateral is 180°.
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