Question

prove that the quadrilateral formed by the internal angle bisectors of any quadrilateral is cyclic

Answer 1


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 + ½ ∠A + ½ ∠ B = 180°
∠AEB  = 180° – ½ (∠A + ∠ B) -------- (2)

From (1) and (2),
∠HEF = 180° – ½ (∠A + ∠ B) --------- (3)

Similarly, ∠HGF = 180° – ½ (∠C + ∠ D) -------- (4)

From 3 and 4,
∠HEF + ∠HGF = 360° – ½ (∠A + ∠B + ∠C + ∠ D)
                          = 360° – ½ (360°)
                          = 360° – 180°
                          = 180°
So, EFGH is a cyclic quadrilateral since the sum of the opposite
angles of the quadrilateral is 180°.]

Answer 2

$\huge\underline\mathbb\blue{ANSWER}$

$\underline\mathcal{GIVEN}$

Let ABCD be a quadrilateral in which the angle bisectors AH, BF, CF and DH of internal angles of A, B, C and D respectively form a quadrilateral EFGH.

$\underline\mathcal{To Prove }$

EFGH is a cyclic quadrilateral.

$\angle{E} + \angle{G} = 180 \degree \: or \: \angle{F} + \angle{H} = 180 \degree$

$\underline\mathcal{Proof }$

In ∆AEB,

$\angle{EAB} + \angle{EBA} + \angle{AEB} = 180 \degree \\ =>\angle{AEB} = 180 \degree - (\angle{EAB} + \angle{EBA})$

Since,

$\angle{FEH } = \angle{ AEB}$

(vertically opposite angles)

$= 180 \degree - ( \angle{EAB} + \angle{EBA}) \\ = 180 \degree - \frac{1}{2} ( \angle{EAB} + \angle{EBA}) \\ = 180 \degree - \frac{1}{2} ( \angle{A} + \angle{B}) .....(i)$

$[ \therefore \: Angles \: A \: and \: B \: are \: bisector \: angles ]$

Similarly,

$= 180 \degree - ( \angle{GCD} + \angle{GDC}) \\ = 180 \degree - \frac{1}{2} ( \angle{2GCD} + \angle{2GDC}) \\ = 180 \degree - \frac{1}{2} ( \angle{C} + \angle{D}) .....(ii)$