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The area of quadrilateral ABCD is 30, e and F are the three equal parts of AB, m and N are the three equal parts of DC. The area of efnm is calculated
∵S（AMD）=1/3 S（ACD）,S(BCF)=1/3 S(ABC)
∴S(AMD)+S(BCF)=1/3*30=10
∴S(AFCM)=20
∵S(AEM)=S(EFM)=1/2* S(AFM),S(CFN)=S(MNF)=1/2* S(CFM)
∴S(EFNM)=1/2*S(AFCM)=10
The diagonal BD of quadrilateral ABCD is bisected by EF, and the area of quadrilateral aecf is 15 square centimeters. Find the area of quadrilateral ABCD
Because s △ AEF = s △ abd / 3
S△CEF=S△CBD/3
S quadrilateral ABCD = 3S quadrilateral aecf
=15 × 3 = 45 (cm2)
In trapezoidal ABCD, ab ∥ CD, ab ﹥ DC, ∠ a = 38 °, B = 52 °, m and N are the midpoint of DC and ab respectively, and Mn = &# 189; (ab-cd)
It is proved that the parallelogram fnmd is obtained by crossing AB with e through D as de ‖ BC and ab with F through D as DF ‖ Mn, so DM = FN, DF = Mn, because de ‖ BC so ∠ AED = ∠ B = 52 ° and ∠ a = 38 ° so ∠ ADB = 180 - ∠ a - ∠ AED = 180-38-52 = 90 ° because AF = an-fn = AB / 2-DM = AB / 2-CD / 2 = (ab-cd) / 2, AE = ab
As shown in the figure, AC is the diagonal of square ABCD, point O is the midpoint of AC, point q is the point on AB, connecting CQ, DP ⊥ CQ at point E, intersecting BC at point P, connecting OP and OQ; verification: (1) △ BCQ ≌ △ CDP; (2) op = OQ
It is proved that the ∵ quadrilateral ABCD is a square, and ∵ B = ∵ PCD = 90 °, BC = CD, (2 points) ∵ DP ⊥ CQ, ∵ 2 + ≌ 1 = 90 °, and ≌ 1 = ≌ 3, (4 points) in △ BCQ and △ CDP, ≌ B = ≌ pcdbc = CD ≌ 1 = ≌ 3
As shown in the figure, AC is the diagonal of square ABCD, point O is the midpoint of AC, point q is the point on AB, connecting CQ, DP ⊥ CQ at point E, intersecting BC at point P, connecting OP and OQ; verification: (1) △ BCQ ≌ △ CDP; (2) op = OQ
It is proved that: the ∵ quadrilateral ABCD is a square, with ∵ B = ∵ PCD = 90 °, BC = CD, (2 points) ∵ DP ⊥ CQ, ∵ 2 + ≁ 1 = 90 °, and ∵ 1 = ≌ 3, (4 points) in △ BCQ and △ CDP,
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