The diagonal lines AC and BD of parallelogram ABCD intersect at point O, and intersect with AB and CD respectively, and then verify with points E and f; OE = of

The diagonal lines AC and BD of parallelogram ABCD intersect at point O, and intersect with AB and CD respectively, and then verify with points E and f; OE = of

Oh, your question is not very clear. If you say that any straight line passes through point O and then intersects AB, CD and F, you can do as I said below
Because the parallelogram ABCD, so ∠ abd = ∠ CDB, Bo = do, because the straight line EF intersects BD at o point. So ∠ EOB = ∠ FOD, corner, corner, so the triangle BOE is equal to the triangle DOF, so OE = of,

In ladder ABCD, ad parallel BC ad = 2 BC = 4 point m is the midpoint of AD, triangle MBC is equilateral triangle
The moving point PQ moves on the line BC and MC respectively, and the angle MPQ = 60 degrees remains unchanged. Let PC be x and MQ be y, the analytic function of Y with respect to X is obtained
In the above question, when the moving point PQ moves to where, the quadrilateral with two vertices in point PM and point ABCD is a parallelogram? And point out the number of qualified parallelogram?
When y is the minimum, the shape of triangle PQC is judged and the reason is explained

As shown in the figure, in the trapezoidal ABCD, ad ‖ BC, ad = 2, BC = 4, point m is the midpoint of AD, and △ MBC is an equilateral triangle. The moving points P and Q move on the line BC and MC respectively (not coincident with the endpoint), and ∠ MPQ = 60 ° remains unchanged

It is known that in ladder ABCD, AD / / is the same as BC (AD)

According to the condition in the question: in ABCD, AD / / in BC (AD)

All the time, in ladder ABCD, AD / / BC (AD / / BC)

1. Extend an BC extension line to f
2. M is the midpoint of AB, and me / / an
3. So e is the midpoint of BF
4. N is the midpoint of AF
5、EN=1/2AB=AM