In the parallelogram ABCD, e is the midpoint of diagonal AC, EF is perpendicular to F, angle B = 60 degrees, ab = 4, angle ACB = 45 degrees, find the length of DF

In the parallelogram ABCD, e is the midpoint of diagonal AC, EF is perpendicular to F, angle B = 60 degrees, ab = 4, angle ACB = 45 degrees, find the length of DF

ABCD is in a parallelogram
∴BC=AD
∠ACB=∠CAD=45°
∵∠B=60°
∴∠BAC=75°
According to the sine theorem
AB/sin45°=AC/sin60°
4/(√2/2)=AC/(√3/2)
AC=2√6
BC/sin75°=AB/sin45°
BC/[(√6+√2)/4]=4/(√2/2)
BC=2√3+2
∵ e is the midpoint of the diagonal AC, EF ⊥ ad
∴AE=1/2AC=√6
The △ AEF is an isosceles right triangle (∠ CAD = 45 °)
∴2AF²=AE² 2AF²=6 AF=√3
∴DF=AD-AF
=BC-AF
=2√3+2-√2
=√3+2

In the parallelogram ABCD, make a straight line AC and ad to OE through point B, and cross the extension line of CD to F, and prove that the square of OB = OE * of

From the similarity of OAB and OCF, ob / of = OA / OC can be obtained. From the similarity of OAE and OBC, ob / OE = OC / OA can be obtained. The two sides of OAE and OBC multiply each other to get the result

In the parallelogram ABCD, the line be intersects AC, ad and the extension line of CD at O, e and F. the following results are obtained: (1) ob square = OE · of (2) AE / ad = CD / CF

(1) Firstly, it is proved that △ AOB is similar to △ COF, then ob: of = OA: OC;
It is proved that △ cob is similar to △ AOE, then OE: OB = OA: OC
So ob: of = OE: OB = OA: OC, so ob square = OE · of
(2) Brief

In the parallelogram ABCD, a straight line AC is made through B, ad is in O, and the extension line of intersection CD is in F. the proof is ob * ob = OE * of

In fact, the triangle AOC is similar to boa, so of: OC = ob: OA, or expressed as of * OA = ob * OC. In addition, the triangle AOE is similar to cob, so OA: OE = OC: OB, or expressed as OA * ob = OC * OE