Parallelogram ABCD parallelogram abef is on the same side AB, m, n are on the diagonal AC, BF respectively, and am: AC = FN: FB to prove Mn / / plane ADF Try to be clear

Parallelogram ABCD parallelogram abef is on the same side AB, m, n are on the diagonal AC, BF respectively, and am: AC = FN: FB to prove Mn / / plane ADF Try to be clear

Make ng ‖ AF intersect AB at point G and link mg
Then FN: FB = Ag: ab = am: AC
∴MG∥BC∥AD
And ∵ mg intersects GN at point G, Da intersects AF at point a
So planar MGN ‖ DAF
So the Mn ‖ plane DAF

As shown in the figure, AC is the diagonal of rectangle ABCD, EF bisects AC at point O, and intersects AD and BC at points E and f respectively

It is proved that the ∵ quadrilateral ABCD is a rectangle, ∵ ad = BC, ad ∥ BC. ∵ CAE = ∵ ACF, ∵ AEF = ∵ CFE. ∵ Ao = Co, ≌ AOE ≌ COF. ∥ AE = CF. ∥ ed = BF

In rectangular ABCD, AC and BD intersect at point O, make ef ⊥ BD when passing through point O, intersect AD and BC at points E and F, and EF = BF, then calculate ∠ Bao

Because EF = BF, fo = 1 / 2ef, so fo = 1 / 2BF, and because the triangle BOF is a right triangle, so the angle oBf is 30 degrees, then the angle Bao is 60 degrees

Known: as shown in the figure, EF is the midpoint of diagonal acbd of quadrilateral ABCD respectively

Proof: & nbsp;
Take the midpoint m of BC and connect EM and FM & nbsp;
Because e is the midpoint of AC and M is the midpoint of BC & nbsp;
So EM is the median of △ ABC & nbsp;
So EM = AB / 2 & nbsp;
Similarly, FM = CD / 2 & nbsp;
Because in △ EFM, there are: EF < EM + FM & nbsp;
So EF < AB / 2 + CD / 2 & nbsp;
That is: EF ＜ (AB + CD) / 2 & nbsp;