In trapezoidal ABCD, AD / / BC, ∠ ABC = 60 °, DCB = 30 °, ab = 4, BC-AD is calculated

In trapezoidal ABCD, AD / / BC, ∠ ABC = 60 °, DCB = 30 °, ab = 4, BC-AD is calculated

For AE ⊥ BC in E, DF ⊥ BC in F,
Because ∠ B = 60 ° AB = 4,
Then be = 2, AE = DF = 2 √ 3,
And ∠ C = 30 ° i.e. FC = 6,
So BC - ad = be + FC = 8

As shown in the figure, in the rectangular ABCD, ab = m (M is a constant greater than 0), BC = 8, e is the moving point on the line BC, connecting De to make ef ⊥ De, where EF intersects the ray ba
For point F, let CE = x, BF = y. if y = 12 / m, what is the value of m to make △ def an isosceles triangle

If △ DEF is isosceles triangle
Then EF = de
Angle Feb + angle Dec = 90 ° angle EDC + angle Dec = 90 °
Therefore, Feb = EDC
And ∠ B = ∠ C
So △ FBE ≌ △ ECD
So EC = FB = y, be = DC = M
BC = be + EC, i.e. m + y = 8
Y = 12 / M
So m equals two or six

As shown in the figure, in the trapezoidal ABCD, ad ‖ BC, e is the midpoint of AB, ∠ EDC = 90 °. Given de = 4, CD = 6, ad = 3, find the length of BC

The extension line of extended de and CB intersects at F
Because ad is parallel to BC
So angle ead = angle EBF
Angle EDA = angle f
Because point E is the midpoint of ab
So AE = be
So triangle ade and triangle BFE are congruent (AAS)
So ad = BF
DE=EF
Because DF = de + ef
DE=4
Angle EDC = 90 degrees
According to Pythagorean theorem, it is concluded that
CF^2=DF^2+CD^2
Because CD = 6
So CF = 10
Because CF = BC + BF
AD=BF=3
So BC = 10-3 = 7

As shown in the figure, in the quadrilateral ABCD, ∠ A and ∠ B complement each other, ∠ C = 90 °, de ⊥ AB, e is the perpendicular foot. If ∠ EDC = 60 °, calculate the degrees of ∠ B, ∠ A and ∠ ade

The ∵ A and ∵ B complement each other, that is ∵ a + ∵ B = 180 °, ∵ ad ∥ BC, ∵ ACD + ∵ ADC = 180 °, and ∵ de ⊥ AB, ∵ ADC = 90 °, ∵ ade = ∵ ADC - ∵ EDC = 90-60 = 30 °, ∵ in the right angle △ AED, ∵ a = 90-30 = 60 °,