Question

lin the adjacent figure ABCD is a square and triangle APB is an equilateral triangle prove that triangle APB is equals to triangle BPC hint in triangle APDand triangle BPC ad = BC ,ap=bp and pad= pbc 90 -60=30​

Answer 1

Δ APD  ≅ Δ BPC  where APB is an equilateral triangle in ABCD square

Step-by-step explanation:

∠BAP = ∠ABP = 60°  ( equilateral Triangle)

∠A = ∠B = 90°

∠DAP = ∠A - ∠BAP = 90° - 60° = 30°

∠CBP = ∠B - ∠ABP = 90° - 60° = 30°

in Δ APD  & Δ BPC

AD = BC  ( Equal Side of Square)

AP = BP   ( Equal  side of Equilateral Triangle)

∠DAP = ∠CBP  = 30°

=> Δ APD  ≅ Δ BPC

QED

Proved

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