
Prove that the quadrilateral formed (if possible) by the internal angle bisectors of any quadrilateral is cyclic.
Answer
585.6k+ views
Hint: A quadrilateral is a polygon having four sides and four vertices. A quadrilateral is cyclic when all of its vertices lie on a circle.
A cyclic quadrilateral is given by the following figure:
A quadrilateral formed by the internal angle bisectors of a quadrilateral ABCD is PQRS:
We need to prove PQRS is cyclic.
A rule of angle equality is, vertically opposite angles are equal.
Angle sum property of a triangle gives the sum of all angles of a triangle = ${180^0}$
To prove the quadrilateral PQRS is cyclic, it is enough to prove that the sum of opposite angles of PQRS is ${180^0}$ . Implies $\angle SPQ + \angle SRQ = {180^0}$
Complete step-by-step answer:
Step 1: Consider a quadrilateral ABCD with internal bisectors AQ, BS, CS, DQ of angles $\angle A$, $\angle B$ , $\angle C$ and $\angle D$ respectively. We need to prove the quadrilateral PQRS formed by these four internal angle bisectors is cyclic.
As vertically opposite angles are equal, we obtain the following equalities:
$
\angle SPQ = \angle APB \\
\angle SRQ = \angle DRC \\
$
Adding above two formulas we get $\angle SPQ + \angle SRQ = \angle APB + \angle DRC$
As AQ is the angle bisector of $\angle A$ and P is a point on AQ, By angle sum property,
$\angle APB = 180 - (\dfrac{1}{2}\angle A + \dfrac{1}{2}\angle B)$ and $\angle DRC = 180 - (\dfrac{1}{2}\angle D + \dfrac{1}{2}\angle C)$ . Thus,
$
\angle SPQ + \angle SRQ = \angle APB + \angle DRC \\
= 180 - (\dfrac{1}{2}\angle A + \dfrac{1}{2}\angle B) + 180 - (\dfrac{1}{2}\angle D + \dfrac{1}{2}\angle C) \\
= 360 - \dfrac{1}{2}(\angle A + \angle B + \angle C + \angle D) \\
= 360 - \dfrac{1}{2}(360) \\
= {180^0} \\
$
The sum of opposite angles of quadrilateral PQRS is ${180^0}$ . Thus PQRS is a cyclic quadrilateral.
Hence proved.
The quadrilateral formed by internal angle bisectors of a quadrilateral is cyclic.
Note: Students should always draw a diagram for better understanding of such questions. Also, they should avoid writing the angles in one letter when taking in consideration a figure where there are several other angles which might seem like the same.
Kite, Trapezoid, Parallelogram, Square, Rhombus, Rectangle comes under Quadrilateral.
A cyclic quadrilateral is given by the following figure:
A quadrilateral formed by the internal angle bisectors of a quadrilateral ABCD is PQRS:
We need to prove PQRS is cyclic.
A rule of angle equality is, vertically opposite angles are equal.
Angle sum property of a triangle gives the sum of all angles of a triangle = ${180^0}$
To prove the quadrilateral PQRS is cyclic, it is enough to prove that the sum of opposite angles of PQRS is ${180^0}$ . Implies $\angle SPQ + \angle SRQ = {180^0}$
Complete step-by-step answer:
Step 1: Consider a quadrilateral ABCD with internal bisectors AQ, BS, CS, DQ of angles $\angle A$, $\angle B$ , $\angle C$ and $\angle D$ respectively. We need to prove the quadrilateral PQRS formed by these four internal angle bisectors is cyclic.
As vertically opposite angles are equal, we obtain the following equalities:
$
\angle SPQ = \angle APB \\
\angle SRQ = \angle DRC \\
$
Adding above two formulas we get $\angle SPQ + \angle SRQ = \angle APB + \angle DRC$
As AQ is the angle bisector of $\angle A$ and P is a point on AQ, By angle sum property,
$\angle APB = 180 - (\dfrac{1}{2}\angle A + \dfrac{1}{2}\angle B)$ and $\angle DRC = 180 - (\dfrac{1}{2}\angle D + \dfrac{1}{2}\angle C)$ . Thus,
$
\angle SPQ + \angle SRQ = \angle APB + \angle DRC \\
= 180 - (\dfrac{1}{2}\angle A + \dfrac{1}{2}\angle B) + 180 - (\dfrac{1}{2}\angle D + \dfrac{1}{2}\angle C) \\
= 360 - \dfrac{1}{2}(\angle A + \angle B + \angle C + \angle D) \\
= 360 - \dfrac{1}{2}(360) \\
= {180^0} \\
$
The sum of opposite angles of quadrilateral PQRS is ${180^0}$ . Thus PQRS is a cyclic quadrilateral.
Hence proved.
The quadrilateral formed by internal angle bisectors of a quadrilateral is cyclic.
Note: Students should always draw a diagram for better understanding of such questions. Also, they should avoid writing the angles in one letter when taking in consideration a figure where there are several other angles which might seem like the same.
Kite, Trapezoid, Parallelogram, Square, Rhombus, Rectangle comes under Quadrilateral.
Recently Updated Pages
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Differentiate between action potential and resting class 12 biology CBSE

Two plane mirrors arranged at right angles to each class 12 physics CBSE

Which of the following molecules is are chiral A I class 12 chemistry CBSE

Name different types of neurons and give one function class 12 biology CBSE

Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Who among the following opened first school for girls class 9 social science CBSE

What does the word meridian mean A New day B Midday class 9 social science CBSE

What is the full form of pH?

Write the 6 fundamental rights of India and explain in detail

Which places in India experience sunrise first and class 9 social science CBSE

