Calculate :

(i) ∠ADC

(ii) ∠ABC

(iii) ∠BAC

(i) Since, DCE is a straight line.

∴ ∠ACD + ∠ACE = 180°

⇒ ∠ACD + 130° = 180°

⇒ ∠ACD = 180° - 130° = 50°.

In △ ADC,

⇒ AD = DC (Given)

⇒ ∠DAC = ∠ACD = 50° (Angles opposite to equal sides are equal.)

By angle sum property of triangle,

⇒ ∠ADC + ∠DAC + ∠ACD = 180°

⇒ ∠ADC + 50° + 50° = 180°

⇒ ∠ADC + 100° = 180°

⇒ ∠ADC = 180° - 100° = 80°.

Hence, ∠ADC = 80°.

(ii) Since, BDC is a straight line.

∴ ∠ADB + ∠ADC = 180°

⇒ ∠ADB + 80° = 180°

⇒ ∠ADB = 180° - 80° = 100°.

In △ ABD,

⇒ AD = BD (Given)

⇒ ∠DBA = ∠BAD = x (let).

By angle sum property of triangle,

⇒ ∠BAD + ∠ADB + ∠DBA = 180°

⇒ x + 100° + x = 180°

⇒ 2x + 100° = 180°

⇒ 2x = 180° - 100°

⇒ 2x = 80°

⇒ x = $\dfrac{80°}{2}$ = 40°.

⇒ ∠DBA = ∠BAD = 40°.

From figure,

⇒ ∠ABC = ∠DBA = 40°.

Hence, ∠ABC = 40°.

(iii) From figure,

⇒ ∠BAC = ∠BAD + ∠DAC = 40° + 50° = 90°.

Hence, ∠BAC = 90°.