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Question
In the following diagram, ABCD is a square and APB is an equilateral triangle.
(i) Prove that: ΔAPD≅ ΔBPC
(ii) Find the angles of ΔDPC.
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Solution
Given: ABCD is a Square and ΔAPB is an equilateral triangle.
We need to
(i) Prove that: ΔAPD≅ ΔBPC
(ii) Find the angles of ΔDPC
Proof:
Since AB side is present in both square & equilateral triangle
AP = PB = AB =AD = CD = BC
(i) In ΔBPC,
BP = BC
∴ ∠BPC = ∠PCB
∠BPC + ∠PCB + 30° = 180°
∠BPC + ∠BPC = 150°
2∠BPC = ` (150°)/2 = 75°`
∴ ∠BPC = ∠PCB = 75°
∠ADP = ∠DPA = 75° ...[C.P.C.T.C]
(ii) In DPC
∠DCP = 90° - 75° = 15°
∠PDC = 90° - 75° = 15°
∠DPC = 180° - (15° + 15°)
∠DPC = 150°
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