# Geometry Sum | ABCD is a square. Prove in each case ∆APD ≅ ∆BPC | Find Angle DPC Angle PCD and Angle PCD |

ABCD is a square and APB is an equilateral triangle. Prove in each case: 1) ∆ APD ≅ ∆ BPC 2) Find the angles of ∆ DPC Solution : First Part Given ∆ ABP = Equilateral triangle R.T.P. 1) ∆ APD ≅ ∆ BPC 2) Find ∠ DPC, ∠ PDC and ∠ PCD, Proof: In ∆ APD and ∆ BPC AP = BP | Same side of the equilateral triangle. AD = BC | Same side of the Square. and ∠ DAP = ∠ DAB - ∠ PAB = 90 ° -60 ° =30 0 Similarly ∠ BPC = ∠ ABC - ∠ ABP = 90 ° -60 ° =30 ° ∴ ∠ DAP = ∠ BPC ∴ ∆ APD ≅ ∆ BPC SAS_Proved 2nd Part In ∆ APD AP=AD || As AP = AB (Equilateral Triangle). We know that ∠ DAP = 30 ° ∴ ∠ APD = (180 ° - 30 ° )/2 || ∆ APD is an isoscless ∆