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CAT 2025 Lesson : Circles - Quadrilaterals in Circles

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7. Inscribed & Circumscribed Quadrilaterals

Properties Figure
Concyclic Property: If a line segment subtends equal angles at its same side, then the end points of the line segment and the vertices of the equal angles are concyclic, i.e., all the points lie on the same circle.

AB subtends equal angles at C and D, i.e., ∠\angleADB =∠= \angleACB.

∴\therefore A, B, C and D are concyclic and lie on the same circle.
Cyclic Quadrilateral: A quadrilateral which can be circumscribed by a circle, i.e., where the 44 vertices are concyclic.

Property 1: The sum of any two opposite angles of a cyclic quadrilateral is 180o180^\mathrm{o}.

∠1+∠3=∠2+∠4=180o\angle1 + \angle3 = \angle2 + \angle4 = 180^\mathrm{o}

Property 2: An exterior angle of a quadrilateral is equal to its interior opposite angle.

∠5=\angle{5} = ∠3\angle{3}, ∠6=\angle{6} = ∠4\angle{4}
∠7=\angle{7} = ∠1\angle{1}, ∠8=\angle{8} = ∠2\angle{2}

Property 3: The product of the length of diagonals is equal to the sum of the product of the lengths of the opposite sides (Ptolemy's Theorem).

AC ×\times BD = AB ×\times CD ++ BC ×\times DA

Property 4: Where a,b,ca, b, c and dd are the lengths of the sides of the cyclic quadrilateral and s=s = a+b+c+d2\dfrac{a + b + c + d}{2}

Area =(s−a)(s−b)(s−c)(s−d)= \sqrt{(s - a)(s - b)(s - c)(s - d)}



Tangential Quadrilateral: The sum of opposite sides of a tangential quadrilateral are equal.

Here, AB ++ CD == BC ++ DA

Example 8

In the diagram below, A, B, C and D lie on a circle. ∠\angleABC == 70o70^\mathrm{o} and ∠\angleBCD == 130o130^\mathrm{o}. BA and CD are extended to points F and E respectively such that ∠\angleFAE == ∠\angleBAD. What is the value of ∠\angleDEA?

Solution

As ABCD is a cyclic quadrilateral,
∠\angleABC =∠= \angleADE =70o= 70^\mathrm{o} (Exterior ∠\angle == Interior Opposite ∠\angle)

∠\angleBAD == 180o180^\mathrm{o} −- ∠\angleBCD == 50o50^\mathrm{o} (Sum of Opposite ∠\angle == 180o180^\mathrm{o})

The question states ∠\angleFAE == ∠\angleBAD == 50o50^\mathrm{o} ∠\angleFAE + ∠\angleDAE ++ ∠\angleBAD == 180o180^\mathrm{o} (∠\angle es on a straight line)
⇒ ∠\angleDAE == 180o−50o−50o180^\mathrm{o} - 50^\mathrm{o} - 50^\mathrm{o} == 80o80^\mathrm{o}

∠\angleAED ++ ∠\angleDAE ++ ∠\angleADE == 180o180^\mathrm{o} (Sum of ∠\angle es of triangle)
⇒ ∠\angleAED ++ 80o+70o=180o80^\mathrm{o} + 70^\mathrm{o} = 180^\mathrm{o}
⇒ ∠\angleAED == 30o30^\mathrm{o}

Answer: 30o30^\mathrm{o}


Example 9

In quadrilateral ABCD, AB, BC, CD and DA are the tangents of a circle with centre O. If AB == 55 cm and BC == 77 cm, then what is the difference in the lengths (in cm) of CD and DA?

Solution

As ABCD is a tangential quadrilateral, the sum of its opposite sides are equal.

∴\therefore AB ++ CD == BC ++ DA
⇒ 55 ++ CD == 77 ++ DA
⇒ DA −- CD == 22

∴\therefore Difference between DA and CD is 22 cm.

Answer: 22


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