In △ABC, angle ∠ACB = 90°. D is a point on side AB so that DA = DC.

(i) Prove that △BDC is an isosceles triangle.

(ii) If ∠BDC = 60°, show that ∠A = 30°.

(i) In △ADC,

Since AD = DC, △ADC is an isosceles triangle.

In isosceles triangle, angles opposite to equal sides are equal.

∴ ∠DAC = ∠DCA.

Let ∠DCA be 'x'.

In △ABC,

∠ACB = 90°

⇒ ∠ACB = ∠DCA + ∠DCB

⇒ ∠DCB = ∠ACB - ∠DCA

⇒ ∠DCB = 90° - x ……..(1)

By angle sum property of triangle,

In triangle ABC,

⇒ ∠ACB + ∠A + ∠B = 180°

⇒ ∠B = 180° - (∠ACB + ∠A)

⇒ ∠B = 180° - (90° + x)

⇒ ∠B = 90° - x ……..(2)

From equation (1) & (2), we get :

∠DCB = ∠B

Since, ∠DCB = ∠B, the sides opposite to them must be equal (CD = BD)

∴ △BDC is an isosceles triangle.

Hence, proved that △BDC is an isosceles triangle.

(ii) Given, ∠BDC = 60°

From figure,

∠ADC + ∠BDC = 180°

∠ADC + 60° = 180°

∠ADC = 120°

Since, triangle ADC is an isosceles triangle, with AD = CD.

Thus, ∠DAC = ∠DCA = a (let)

By angle sum property of triangle,

∠DAC + ∠DCA + ∠ADC = 180°

a + a + 120° = 180°

2a = 180° - 120°

2a = 60°

a = $\dfrac{60\degree}{2}$ = 30°.

Hence proved that ∠A = 30°.