Question

ABCD is a cyclic quadrilateral whose diagonal intersect at point E. If <DBC=70°, <BAC=30°, find <BCD. Further, if AB=BC, find<ECD​

Answer 1

Given :

• ABCD is a cyclic quaderilateral
• ∠DBC = 70°
• ∠BAC = 30°
• AB = BC

To find :

• ∠ECD
• ∠BCD

According to the question,

⇒ ∠BDC = ∠BAC

⇒ ∠BAC = 30°

Reason : Angle in the same segment

In △BCD,

⇒ ∠BCD + ∠DBC + ∠BDC = 180°

⇒ ∠BCD + 70° + 30° = 180°

⇒ ∠BCD + 100° = 180°

⇒ ∠BCD = 180° - 100°

⇒ ∠BCD = 80°

⇒ ∠BAC = ∠BCA

⇒∠BCA = 30°

Reason : Radius

⇒ ∠ECD = ∠BCD - ∠BCA

⇒ ∠ECD = 80° - 30°

⇒ ∠ECD = 50°

So,

• ∠BCD = 80°
• ∠ECD = 50°

___________________

Theorem 1 :

The angle subtended by an arc of a circle at the centre is doubled the angle subtended by it at any point on the remaining part of the circle.

Theorem 2 :

The angle in a semicircle is a right angle.

Theorem 3 :

Angles in the same segment of a circle are equal.

Answer 2

✦Given:-

⮕ABCD is a Cyclic quadrilateral

⮕Diagonals of quadrilateral intersect at point E

⮕∠ DBC = 70°

⮕∠ BAC = 30°

⮕AB = BC

✦To Find:-

⮕∠ BCD

⮕∠ ECD

✦Solution:-

For Chord CD,

⮕∠ CBD = ∠ CAD

⮕∠ CAD = 70°

⮕∠ BAD = ∠ BAC + ∠ CAD

⮕∠ BAD = 30 + 70

⮕∠ BAD = 100°

⮕∠ BCD + ∠ BAD = 180° ( PROPERTY OF CYCLIC QUAD)

⮕∠ BCD + 100 = 180°

⮕∠ BCD = 180 - 100

⮕∠ BCD = 80°

⮕In Triangle ABC

⮕AB = BC

⮕∠ BCA = ∠ CAB

⮕∠ BCD = 80°

⮕∠ BCA + ∠ BCD = 80°

⮕30° + ∠ BCD = 80°

⮕∠ BCD = 80 - 30

⮕∠ BCD = 50°

⮕∠ BCD = ∠ ACD = ∠ ECD = 50°

⮕Hence, ∠ BCD = 80° , ∠ ECD = 50°.