Question

In Fig. 10.39, A, B, C and D are four points on a circle. AC and BD intersect at a point E such
that < BEC = 130° and < ECD = 20°. Find <BAC.​

Answer 1

Given :

• ∠ECD = 20°
• ∠BEC = 130°

To find :

• ∠BAC

According to the question,

➞ ∠BEC + ∠DEC = 180°

➞ 130° + ∠DEC = 180°

➞ ∠DEC = 180° - 130°

➞ ∠DEC = 50°

Now,

In △ DEC,

➞ ∠DEC + ∠CDE + ∠DCE = 180°

Reason : Angle sum property

➞ 50° + ∠CDE + 20° = 180°

➞ 70° + ∠CDE = 180°

➞ ∠CDE = 180° - 70°

➞ ∠CDE = 110°

So,

➞ ∠CDE = ∠BAC

➞ ∠BAC = 110°

Reason : Angle in the same segment are equal.

___________________

★ More theorem :

★ Theorem 1 :

Angle subtended by an arc at the centre of the circle is doubled the angle made by the same arc at the circumference of the circle.

★ Theorem 2 :

Angle in the semicircle is right angle (90°).

Answer 2

Given :

∠ECD = 20°

∠BEC = 130°

To find :

∠BAC

According to the question,

➞ ∠BEC + ∠DEC = 180°

➞ 130° + ∠DEC = 180°

➞ ∠DEC = 180° - 130°

➞ ∠DEC = 50°

Now,

In △ DEC,

➞ ∠DEC + ∠CDE + ∠DCE = 180°

Reason : Angle sum property

➞ 50° + ∠CDE + 20° = 180°

➞ 70° + ∠CDE = 180°

➞ ∠CDE = 180° - 70°

➞ ∠CDE = 110°

So,

➞ ∠CDE = ∠BAC

➞ ∠BAC = 110°

Reason : Angle in the same segment are equal.

___________________

★ More theorem :

★ Theorem 1 :

Angle subtended by an arc at the centre of the circle is doubled the angle made by the same arc at the circumference of the circle.

★ Theorem 2 :

Angle in the semicircle is right angle (90°).