例1如图1,四边形ABCD是边长为1的正方形,△PBC是等边三角形,求△BPD的面积.解如图1,从点D作DE⊥PC于点E,从点P作PF⊥BC于点F.因为正方形ABCD的边长是1,所以S_(△BCD)=1/2,因为△PBC是等边三角形,所以BC=PB=PC=CD=1,∠PBC=∠PCB=60°,于是PF=(3~(1/2))/2,∠DCE=30°, Example 1 As shown in Figure 1, the quadrilateral ABCD is a square with a side length of 1. The △ PBC is an equilateral triangle, and the area of ​​△ BPD is obtained. Solution As shown in Figure 1, from point D to DE⊥PC at point E, ⊥BC at point F. Since the side length of the square ABCD is 1, S_ (ΔBCD) = 1/2, BC = PB = PC = CD = 1, ∠PBC = ∠PCB because ΔPBC is an equilateral triangle = 60 °, then PF = (3 ~ (1/2)) / 2, ∠DCE = 30 °,