Test: Cyclic Quadrilaterals - Grade 9 MCQ

We know that, sum of all angles of rectangle =360° so, the sum of 2 angles of rectangle = 360°\2 = 180°  Angle A + Angle B=180°

  75+B=180°

or, B=180-75°

or, B=105 degree.

Using the cyclic quadrilateral property:

∠BEC + ∠BAC = 180^∘

Substitute the value of ∠BEC=140^∘:

∠BAC = 180^∘−140^∘= 40^∘

∠ECD is an exterior angle to triangle △CED.

• By the Exterior Angle Property:

∠ BEC = ∠BAC + ∠ECD

140^∘=∠BAC+25^∘

∠BAC = 140^∘−25^∘=115^∘

∵∠BAC and ∠BDC are in the same segment.

∴ ∠BAC=∠BDC=45^∘

Now in ΔBCD, ∠DBC+∠BDC+∠BCD=180^∘ (Sum of angles of a triangle)

⇒55^∘+45^∘+∠BCD=180^∘

⇒100^∘+∠BCD=180^∘

⇒∠BCD=180^∘−100^∘ = 80^∘

Hence, ∠BCD=80^∘

Property of Cyclic Quadrilateral: In a cyclic quadrilateral, the sum of opposite angles is always 180°:

∠P + ∠R = 180^∘

Given ∠P = 80^∘, substitute into the equation:

80^∘+∠R =180^∘

Rearranging the equation:

∠R = 180^∘−80^∘=100^∘

The sum of opposite angles of a cyclic quadrilateral is 180°.

M + O=180°

O=180°- 82°= 98°

We know that ABCD is a cyclic quadrilateral

So we get

∠ABC + ∠ADC = 180^o

By substituting the values

92^o + ∠ADC = 180^o

On further calculation ∠ADC = 180^o – 92^o

By subtraction ∠ADC = 88^o

We know that AE || CD

From the figure we know that

∠EAD = ∠ADC = 88^o

We know that the exterior angle of a cyclic quadrilateral = interior opposite angle

So we get

∠BCD = ∠DAF

We know that

∠BCD = ∠EAD + ∠EAF

It is given that ∠FAE = 20^o

By substituting the values

∠BCD = 88^o + 20^o

By addition ∠BCD = 108^o

Therefore, ∠BCD = 108^o.

Definition of Concyclic Points: Four points are said to be concyclic if they lie on the circumference of a single circle. In such a case, the opposite angles of the quadrilateral formed by these points are supplementary:

∠C+∠A=180^∘ and ∠D+∠B=180^∘

Analyze ∠C+∠A

From the figure, ∠A is the angle opposite to ∠C. The condition for cyclic quadrilaterals is:

∠C+∠A=180^∘

Substitute ∠C=50^∘:

50^∘+∠A=180^∘

Solve for ∠A:

∠A = 180^∘− 50^∘=130^∘

 Analyze ∠D+∠B

Similarly, ∠B is opposite to ∠D. The condition for cyclic quadrilaterals is:

∠D+∠B = 180^∘

Substitute ∠D=50^∘

50^∘+∠B = 180^∘

Solve for ∠B:

∠B=180^∘−50^∘=130^∘

 Check Concyclicity

• Opposite angles ∠C+∠A=180^∘
• Opposite angles ∠D+∠B=180^∘

Since the opposite angles of quadrilateral ABCD are supplementary, the quadrilateral is cyclic. Therefore, the four points A,B,C,D are concyclic and lie on the circumference of a single circle.

We know that if one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior
opposite angle.
i.e., ∠BAD = ∠DCF = 75°
⇒ ∠DCF = x = 75 °
Again, the sum of opposite angles in a cyclic quadrilateral is 180 °.
Thus, ∠DCF + ∠DEF = 180°
⇒ 75° + y = 180°
⇒ y = (180° - 75°) = 105°
Hence, x = 75° and y = 105°

A quadrilateral is called cyclic if all the four vertices of it lie on a circle and the sum of either pair of opposite angles of a cyclic quadrilateral is 180^∘.

Here,∠1+∠2=180^∘