The diagonals AC and BD of parallelogram ABCD intersect at O, e and F, which are two points on DB, and de = BF I'm sorry to trouble you to copy it,

The diagonals AC and BD of parallelogram ABCD intersect at O, e and F, which are two points on DB, and de = BF I'm sorry to trouble you to copy it,

Well, from de = BF, ABCD as parallelogram, ab = CD, angle abd = angle BDC, we can get triangle AEB ≌ triangle DFC, AE = cf. similarly, we can get triangle AFD, congruent triangle BEC, EC = AF, so we can get quadrilateral afce as parallelogram
Tired to death

Given that E and F are two points on the diagonal BD of the parallelogram ABCD, and ED = BF, try to judge whether the parallelogram afce is parallel

Connecting AE, CF, AF, EC
Through SAS, △ Abf ≌ DCE, △ DEA ≌ BFC were proved
That is, AFB = Dec, ∠ BFC = AED
It can be proved that two sides are parallel

As shown in the figure, in the parallelogram ABCD, the points E and F are on the diagonal BD, and de = BF

Analysis: connect AC, intersect BD at a point O, use two groups of diagonals to judge the quadrilateral aecf is a parallelogram. Prove: connect AC, intersect BD at point O, ∵ quadrilateral ABCD is a parallelogram, ∵ Ao = Co, Bo = do ∵ BF = De, ∵ bo-bf = do-de, that is: of = OE, ∵ quadrilateral aecf is a parallelogram. Please click "adopt as the answer"

In the triangle ABC, AB is equal to AC, BD and CE are the heights of the triangle ABC

Your question should be
Verification: quadrilateral ebcd is isosceles trapezoid
∵AB=AC
∴∠EBC=∠DCB
And ∠ BEC = ∠ CDB = 90 ° BC = CB
∴△BEC≌△CDB
∴BE=CD
∴AB-BE=AC-CD
AE = ad
∴∠AED=∠ADE
∴∠AED=∠ADE=∠B=∠C
∴ED//BC
The quadrilateral ebcd is an isosceles trapezoid