Question

prove that the bisector of the angles of parallelogram enclose a rectangle

Answer 1

as give in the question just solve
if the solution is wrong then inform me

Answer 2

$\mathbb{ELLO \ THERE!}$

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Given,

ABCD is a parallelogram.

All the angles in it are bisected.

Bisectors form a quadrilateral.

To Prove,

PQRS is a Rectangle

Proof,

In ABCD

∠A + ∠B = 180° (co - interior angles)

$\frac{1}{2}$∠A + $\frac{1}{2}$∠B = $\frac{1}{2}$ 180°

(Halves of equals are equal)

∴ ∠1 + ∠2 = 90°

In Δ AQB

∠1 + ∠2 + ∠AQB = 180° (A.S.P of Triangles)

90° + ∠AQB = 180°    ( ∠1 + ∠2 = 180°)

∠AQB = 180° - 90°

∠AQB = 90° → 1

Similarly,

∠DSC = 90° → 2

In ABCD,

∠A + ∠D = 180° (co - interior angles)

$\frac{1}{2}$∠A + $\frac{1}{2}$∠D = $\frac{1}{2}$ 180°

(Halves of equals are equal)

∴ ∠5 + ∠6 = 90°

In Δ DPA

∠5 + ∠6 + ∠DPA = 180° (A.S.P of Triangles)

90° + ∠DPA = 180°    ( ∠5 + ∠6 = 180°)

∠DPA = 180° - 90°

∠DPA = 90°

∠DPA = ∠QPS (V.O.A)

∴ ∠QPS = 90° → 3

Similarly,

∴ ∠QRS = 90° → 4

Now,

From 1, 2, 3 and 4

PQRS is a rectangle.

(all angles are 90°)

$\large\boxed{\large\boxed{\mathbb{PROVED!}}}$

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$\textsc{Hope \ It \ Helps}$