As shown in the figure, ad ∥ BC in ladder ABCD, point E is the midpoint of edge ad, connecting be with AC at point F, the extension line of be with CD at point G, verify Ge / GB = AE / BC

As shown in the figure, ad ∥ BC in ladder ABCD, point E is the midpoint of edge ad, connecting be with AC at point F, the extension line of be with CD at point G, verify Ge / GB = AE / BC

It is proved that the triangle GED is similar to the triangle GBC because of ED ‖ BC
Then Ge / GB = ed / BC
Point E is the midpoint of edge ad, that is AE = ed
So Ge / GB = AE / BC

As shown in the figure, in ladder ABCD, ab ‖ DC, ∠ B = 90 °, e is the point on BC, and AE ⊥ ed. if BC = 12, DC = 7, be: EC = 1:2, find the length of ab

∫ ab ∥ DC, and ∠ B = 90 °, ∧ AEB + ∠ BAE = 90 ° and ∠ C = 90 degree. (1 point) ∧ AEB + ∠ CED = 90 degree. So ∠ BAE = ∠ CED. (2 points) ∧ EAB ∧ Dec. (3 points) ABEC = becd. (3 points) be: EC = 1:2, BC = 12 and DC = 7, so ab8 = 47. (4 points) AB = 327. (5 points)

As shown in the figure, the quadrilateral ABCD is an isosceles trapezoid, ab = DC, the diagonal AC = BC = 2Ab, passing through point a to make AE / / DC, intersecting with point e to find the value of be: EC
KUAI

∵AE∥DC
∴AE=DC=AB
Ψ△ Abe isosceles
And AC = BC = 2Ab
Ψ△ cab isosceles
And ∠ B is the common base angle of two isosceles triangles
∴△ABE∽△CAB
And ab = 2be
∴BC=4BE
∴BE:EC=1:3

As shown in the figure, in ladder ABCD, ab ‖ DC, ∠ B = 90 °, e is the point on BC, and AE ⊥ ed. if BC = 12, DC = 7, be: EC = 1:2, find the length of ab

∫ ab ∥ DC, and ∠ B = 90 °, ∧ AEB + ∠ BAE = 90 ° and ∠ C = 90 degree. (1 point) ∧ AEB + ∠ CED = 90 degree. So ∠ BAE = ∠ CED. (2 points) ∧ EAB ∧ Dec. (3 points) ABEC = becd. (3 points) be: EC = 1:2, BC = 12 and DC = 7, so ab8 = 47. (4 points) AB = 327. (5 points)

In the parallelogram ABCD, the diagonals AC and BD intersect at O, EF passes through O, and AF ⊥ BC. It is proved that the quadrilateral afce is a rectangle

It is proved that: ∵ quadrilateral ABCD is parallelogram, ∥ OA = OC, AE ∥ FC, ∥ EAO = ∥ FCO, in △ AOE and △ COF, ∥ EAO = ∥ fcoao = Co ∥ AOE = ∥ COF, ∥ AE = CF, ∥ quadrilateral aecf is parallelogram, and ∥ AF ⊥ BC, ∥ AFC = 90 °, then quadrilateral aecf is rectangle

As shown in the figure, in the parallelogram ABCD, O is the intersection of diagonal AC and BD, through o make ef ⊥ ad to e, intersect BC to F, connect AF and CE. Try to explain: the quadrilateral afce is a parallelogram

It is proved that: as shown in the figure, ∵ quadrilateral ABCD is a parallelogram, ∵ Ao = Co, ad ∥ BC, then AE ∥ FC. And ∵ EF ⊥ ad, ≁ EF ⊥ BC, ∵ AEO = ∠ CFO = 90 ° in △ AOE and △ COF, ≌ AEO = ∠ CFO ∠ AOE = ∠ cofao = Co, ≌ AOE ≌ COF (AAS), ≌ AE = CF, ≌ quadrilateral afce is a parallelogram

As shown in the figure, the diagonals BD and AC of the parallelogram ABCD intersect at O, e, F and G, which are respectively the midpoint of ob, OC and ad, and AC = 2Ab, eg = EF

Proof: let the midpoint of OD be m connection MG.MF
Because F, m and G are OC, OD and ad respectively
So FM and GM are the median lines of triangle ODC and triangle DOA
So GM = 1 / 2oa, FM = 1 / 2CD, OA = CD = ab
So GM = FM
Because FM parallel CD GM parallel OA is equal according to the internal stagger angle, angle GME = angle FME
So triangle EGM and triangle EFM are congruent, so EF = eg

When the diagonals of the parallelogram ABCD intersect at the midpoint of the points O.E, F and P ob, OC and ad respectively, and AC = 2Ab, we prove that EP = EF

It is proved that: connecting AE, ∵ quadrilateral ABCD is parallelogram, ∵ ad = BC, AC = 2oa = 2oC, ∵ AC = 2Ab, ∵ OA = AB, ∵ e is ob midpoint, ∵ AE ⊥ BD (three line in one theorem), ∵ AED = 90 °, ∵ P is ad midpoint, ∵ ad = 2ep, ∵ BC = ad, ∵ BC = 2ep, ∵ E and F are ob midpoint, OC midpoint, ∵ BC = 2ef, ∵ EP = EF respectively

The right triangle CEF is inscribed in the square ABCD, e is the midpoint of AD, and EF is perpendicular to EC. If AB = 1, then the length of CF is?
RT

Let AF be x.fb = 1-x, AE = 0.5, ed = 0.5
X^2+0.25+1.25=(1-X)^2+1
X=0.25
CF = 0.75 ^ 2 + 1 under root sign

As shown in the figure, in a square ABCD, point E is on the side of AD, and AE = quarter ad, f is the midpoint of ab. we prove that △ CEF is a right triangle

Let the side length of a square be x, then AE = 0.25x; de = 0.75x; AF = 0.5x; BF = 0.5x; BC = x; CD = X. in triangle AEF, EF square = 0.25x square + 0.5x square = 0.3125x square, in triangle BCF, FC square = 0.5x square + x square = 1.25x square, in triangle CDE, EC square = 0.75x square + x square = 1