Question

In the following diagram, ABCD is a square and APB is an equilateral triangle.

1. Prove that: ΔAPD ≅ ΔBPC
2. Find the angles of ΔDPC.

Answer 1

Given: ABCD is a Square and ΔAPB is an equilateral triangle.

(i) Proof: In ΔAPB,

AP = PB = AB            ...[APB is an equilateral triangle]

Also, we have,

∠PBA = ∠PAB = ∠APB = 60°        ...(1)

Since ABCD is a square, we have

∠A = ∠B = ∠C = ∠D = 90°             ...(2)

Since ∠DAP = ∠A + ∠PAB        ...(3)

⇒ ∠DAP = 90° + 60°  

⇒ ∠DAP = 150°     ...[From (1) and (2)]       ...(4)

Similarly ∠CBP = ∠B + ∠PBA

⇒ ∠CBP = 90° + 60° 

⇒ ∠CBP = 150°       ...[From (1) and (2)]  ...(5)

⇒ ∠DAP = ∠CBP    ...[From (1) and (2)]  ...(6)

In ΔAPD and ΔBPC

AD = BC          ...[Sides of square ABCD]

∠DAP = ∠CBP   ...[From (6)]

AP = BP            ...[Sides of equilateral ΔAPB]

∴ By Side-Angel-Side Criterion of Congruence, we have,

ΔAPD ≅ ΔBPC

(ii)

AP = PB = AB       ...[ΔAPB is an equilateral triangle] ...(7)

AB = BC = CD = DA   ...[Sides of square ABCD] ...(8)

From (7) and (8), we have

AP = DA and PB = BC                                  ...(9)

In ΔAPD,

AP = DA                    ...[From (9)]

∠ADP = ∠APD   ...[Angel opposite to equal sides are equal]        ...(10)

∠ADP + ∠APD + ∠DAP = 180°  ...[Sum of angel of a triangle = 180°]

⇒ ∠ADP + ∠ADP + 150° = 180°    ...[From (3), ∠DAP = 150° From (10), ∠ADP = ∠APD]

⇒ ∠ADP + ∠ADP = 180° - 150°

⇒ 2∠ADP = 30°

⇒ ∠ADP = `30^circ/2`

⇒∠ADP= 15°

We have ∠PDC =∠D - ∠ADP 

⇒ ∠PDC = 90° - 15°

⇒ ∠PDC = 75°             ...(11)

In ΔBPC, 

PB = BC     ...[From (9)]

∴ ∠PCB = ∠BPC   ...[Angel opposite to equal sides are equal]    ...(12) 

∠PCB + ∠BPC + ∠CBP = 180°    ...[Sum of angel of a triangle = 180°]

⇒ ∠PCB + ∠PCB + 30° = 180°   ...[From (5), ∠CBP = 150° from (12), ∠PCB = ∠BPC]

⇒ 2∠PCB =  180° - 150°

⇒ 2∠PCB = `30^circ/2`

⇒ ∠PCB = 15°

We have ∠PCD = ∠C - ∠PCB 

⇒ ∠PCD = 90° - 15°

⇒ ∠PCD = 75°           ...(13)

In ΔDPC, 

∠PDC = 75° 

∠PCD = 75°

∠PCD + ∠PDC + ∠DPC = 180°   ...[Sum of angles of a triangle = 180°]

⇒ 75° + 75° + ∠DPC = 180°

⇒ ∠DPC = 180° - 150°

⇒ ∠DPC = 30° 

∴ Angles of DPC, are: 75°, 30°, 75°