In the given figure, ABCD is a square and △PAB is an equilateral triangle.

(i) Prove that △APD ≅ △BPC.

(ii) Show that ∠DPC = 15°.

(i) We know that,

Each interior angle in a square is 90° and each interior angle in an equilateral triangle is 60°.

From figure,

⇒ ∠DAP = ∠DAB + ∠BAP

⇒ ∠DAP = 90° + 60° = 150°

⇒ ∠CBP = ∠CBA + ∠ABP

⇒ ∠CBP = 90° + 60° = 150°

In △APD and △BPC,

⇒ ∠DAP = ∠CBP [Each equal to 150°]

⇒ AD = BC [Sides of a square]

⇒ AP = BP [Sides of an equilateral triangle]

∴ △APD ≅ △BPC (By S.A.S axiom)

Hence, △APD ≅ △BPC.

(ii) ABCD is a square.

∴ AB = AD = DC = BC

APB is an equilateral triangle.

∴ AP = PB = AB

So, we get :

AP = AD and PB = BC

∴ △APD and △BPC are isosceles triangle.

We know that,

Angles opposite to equal sides are equal.

∴ ∠APD = ∠ADP = x (let) and ∠BPC = ∠BCP = y (let)

In △APD,

By angle sum property of triangle,

⇒ ∠APD + ∠ADP + ∠DAP = 180°

⇒ x + x + 150° = 180°

⇒ 2x = 180° - 150°

⇒ 2x = 30°

⇒ x = $\dfrac{30°}{2}$

⇒ x = 15°.

In △BPC,

By angle sum property of triangle,

⇒ ∠BPC + ∠BCP + ∠PBC = 180°

⇒ y + y + 150° = 180°

⇒ 2y = 180° - 150°

⇒ 2y = 30°

⇒ y = $\dfrac{30°}{2}$

⇒ y = 15°

From figure,

⇒ ∠PDC = ∠ADC - ∠ADP = 90° - 15° = 75°

⇒ ∠PCD = ∠BCD - ∠BCP = 90° - 15° = 75°

In △DPC,

By angle sum property of triangle,

⇒ ∠DPC + ∠PDC + ∠PCD = 180°

⇒ ∠DPC + 75° + 75° = 180°

⇒ ∠DPC + 150° = 180°

⇒ ∠DPC = 180° - 150°

⇒ ∠DPC = 30°.

Hence, proved that ∠DPC = 30°.