數學4條難題

in triangle EFG

angle FEB = b + c(三角形外角和)

延長FE至AB,設該點是Q

IN triangle QEB

angle BQE = 180 - a - (b + c) (三角形內角和)

= 180 - a - b - c

= d

so,AB//CD


2)angle BOD

= 180 - z - y(直線上的鄰角)

= x

x = 180 - z - y(對頂角)

x + y + z = 180


3)angle BFG = angle DGH = b(同位角,AB//CD)

a + b = 180


4)延長AE及CD相交於K

angle EKD = q - r(三角形外角和)

since AB//CD(GIVEN)

so,
180 - p = q - r(同旁內角AB//CD)

p + q = r + 180

Angle E=b+c (angle sum of triangle) (vert. oppo. angle)
because given a+b+c+d=180
draw a //line through E called EF

CD//EF (alt angle supp.)
EF//AB (alt. angle supp.)

therefore AB//CD