Question

prove that the quadrilateral formed by bisectors of any quadrilateral is a cyclic quadrilateral.

Answer 1

.... Hope you got it.

Answer 2

Question

Prove that the quadrilateral formed by bisectors of any quadrilateral is a cyclic quadrilateral.

Answer

What is a cyclic quadrilateral?

A quadrilateral where all it's vertices ae inscribed in a circle is called a Cyclic Quadrilateral.

To show that any quadrilateral is cyclic, we need to show that the sum of the opposite angles of the desired quadrilateral is equal to 180°

Given

ABCD is a quadrilateral

To Prove

EFGH is a cyclic Quadrilateral.

i.e, ∠FEH + ∠FGH = 180°

[FEH and FGH are opposite angles.]

Proof

Consider ΔAEB

∠AEB = 180° - [∠EAB + ∠EBA]         [A.S.P of a Δ]

∠AEB = 180° - [$\frac{1}{2}$ ∠A + $\frac{1}{2}$ ∠B]

∠AEB = 180° - $\frac{1}{2}$ [∠A + ∠B]

But ∠AEB = ∠FEH

∴ ∠FEH = 180° - $\frac{1}{2}$ [∠A + ∠B]  → ${\large\boxed{\mathsf{1}}}$

Similarly,

∠FGH = 180° - $\frac{1}{2}$ [∠D + ∠C]  → ${\large\boxed{\mathsf{2}}}$

By adding 1 and 2,

∠FEH + ∠FGH = 180° - $\frac{1}{2}$ [∠A + ∠B] + 180° - $\frac{1}{2}$ [∠D + ∠C]

[On adding Numbers, and taking out common factors we get ↓]

∠FEH + ∠FGH = 360° - $\frac{1}{2}$ [∠A + ∠B + ∠D + ∠C]

[ABCD is a quadrilateral, therefore ∠A + ∠B + ∠D + ∠C = 360°]

∠FEH + ∠FGH = 360° - $\frac{1}{2}$ × 360°

∠FEH + ∠FGH = 360° - 180°

∠FEH + ∠FGH = 180°

Since the sum of supplementary angles is 180°,

EFGH is a cyclic quadrilateral.