Prove that the quadrilateral formed (if possible) by the internal angle bisectors of any
quadrilateral is cyclic.
Answers
Step-by-step explanation:
that the quadrilateral formed (if possible) by the internal angle bisectors of any
quadrilateral is cyclic.
Answer:
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 [Vertically opposite angles] -------- (1)
Consider triangle AEB,
∠AEB +
2
1
∠A +
2
1
∠ B = 180°
∠AEB = 180° –
2
1
(∠A + ∠ B) -------- (2)
From (1) and (2),
∠HEF = 180° –
2
1
(∠A + ∠ B) --------- (3)
Similarly, ∠HGF = 180° –
2
1
(∠C + ∠ D) -------- (4)
From 3 and 4,
∠HEF + ∠HGF = 360° –
2
1
(∠A + ∠B + ∠C + ∠ D)
= 360° –
2
1
(360°)
= 360° – 180°
= 180°
So, EFGH is a cyclic quadrilateral since the sum of the opposite
angles of the quadrilateral is 180°.]
solution