In the adjacent figure ABCD is a square and AAPB is anequilateral triangle. Prove that AAPD: ABPC.(Hint: In AAPD and ABPC AD=BC, AP = BP andPAD = Z PBC = 90° -60° - 30)

In the adjacent figure ABCD is a square and AAPB is anequilateral tria

1.	In the adjacent figure ABCD is a square and AAPB is anequilateral triangle. Prove that AAPD: ABPC.(Hint: In AAPD and ABPC AD=BC, AP = BP andPAD = Z PBC = 90° -60° - 30)
Answer» Solution : part 1GivenΔ ABP is an equilateral triangle.Then, AP = BP (Same side of the triangle)and AD = BC (Same side of square)and,∠ DAP =∠ DAB -∠ PAB = 90° - 60° = 30°Similarly,∠ BPC =∠ ABC -∠ ABP = 90° - 60° = 30°∴∠ DAP =∠ BPC∴Δ APD is congruent toΔ BPC ( SAS proved)2nd partInΔ APDAP = AD II as AP = AB (equilateral triangle)We know that∠ DAP = 30°∴∠ APD = (180° - 30°)/2 (Δ APD is an isosceles triangle and∠ APD is on of the base angles.)= 150°/2 = 75°=∠ APD = 75°Similarly,∠ BPC = 75°Therefore,∠ DPC = 360° - (75°+75°+60°)=∠ DPC = 150°Now inΔ PDCPD = PC asΔ APD is congruent toΔ BPC∴Δ PDC is an isosceles triangleAnd∠ PDC =∠ PCD = (180° - 150°)/2or∠ PDC =∠ PCD = 15°∠ DPC = 150°;∠ PDC = 15° and∠ PCD = 15° Answer.