The circumference of rectangle ABCD is 28cm, its two diagonals intersect at point O, the circumference of △ AOB is 2cm shorter than that of △ BOC, then AB = (), BC = ()

The circumference of rectangle ABCD is 28cm, its two diagonals intersect at point O, the circumference of △ AOB is 2cm shorter than that of △ BOC, then AB = (), BC = ()

The perimeter of △ AOB = 2 × half diagonal + ab
The perimeter of △ BOC = 2 × half diagonal + BC
The perimeter of △ AOB is 2cm shorter than that of △ BOC, and ab is 2cm shorter than BC
AB+BC=28÷2=14
AB = (14-2) △ 2 = 6cm
BC = 6 + 2 = 8 cm

It is known that a is outside the plane BDC, ab = AC = BC = ad = CD = dB, e is the midpoint of AD. find the angle between CE and plane BCD

Let f be the midpoint of BC and g the perpendicular foot of E on BCD
sin∠EFD＝（1/2）／（√3/2）＝1/√3.
cos∠EFD＝√（2/3）.
EF＝FD×cos∠EFD＝（√3／2）×√（2/3）＝1/√2.
FG＝FE×cos∠EFD＝（1/√2）×√（2/3）＝1/√3.
CG²＝CF²＋FG²＝（1/2）²＋（1/√3）²＝7/12.
CG＝√（7/12）.
cos∠ECG＝CG／CE＝√（7/12）/（√3／2）＝√7／3.
∠ECG＝arccos（√7/3）.
This is the angle between CE and BCD

As shown in the figure, let a be the point out of the plane where BCD is located, ad = BC = 2cm, e and f be the midpoint of AB and CD respectively. If EF = root 2cm, find the intersection of AD and BC

The right angle takes the midpoint g of BD to connect eg, FG, eg = FG = 1, EF = 2, so the triangle EFG is a right triangle
Eg / / AD, FG / / BC so the angle formed by ADB is right angle

As shown in the figure, in the right triangle ABC, ∠ C = 90 ° and BC is bisected by points D and E, and BC = 3aC, then ∠ AEC + ∠ ADC + ∠ ABC=______ °．

As shown in the figure, bcmn is a square with BC as the edge, and points P and Q are taken on Mn to make MP = PQ = QN. If AP, PD and DQ are connected, then am = CD, MP = AC, △ ACB ≌ △ PQD can be obtained easily, so AP = ad and ∠ DAC + ∠ PAM = 90 °, i.e. △ APD is an isosceles right triangle, so ∠ ABC + ∠ ADC = ∠ PDQ + ∠ ADC = 90 ° - ADP = 90 ° - 45 ° = 45 ° and ∠ AEC = ∠ EAC = 45 ° so ∠ AEC + ∠ ADC + ∠ ABC = 9 0°