In circle O, radius r = 5cm, chord ad// BC.AD=6cm BC = 8cm (1) what kind of special quadrilateral ABCD is? Prove your conjecture In circle O, radius r = 5cm, chord ad// BC.AD=6cm BC=8cm (1) What kind of special quadrilateral is quadrilateral ABCD? Prove your conjecture; (2) Find the area of quadrilateral ABCD

Proof: connect AC
∵AB∥CD.
∴∠DCA=∠BAC.
The arc ad = arc BC
∴AD=BC.
So the trapezoid ABCD is isosceles trapezoid
I'm also writing this problem. I found it only when I found it dead. I can understand it by looking at it. There is a picture. Please think about the second question

In the isosceles trapezoid ABCD, ad is parallel to BC, BD is vertical to CD, and the angle ABC = 60 ° BC = 16cm

Because the quadrilateral ABCD is isosceles trapezoid
So ∠ ABC = ∠ C, ab = DC
Because ∠ ABC = 60 degrees
So ∠ C = 60 degrees
Because BD ⊥ CD, ∠ C = 60 ° BC = 16 cm
So CD = 8 cm
So AB = 8 cm
Make a height at point a and point d respectively
Because ∠ ABC = ∠ C = 60 ° CD = AB = 8 cm
So ad = 16-8 △ 2 × 2 = 8 cm
So perimeter = 16 + 8 + 8 + 8 = 40 cm

As shown in the figure, the circumference of diamond ABCD is 16cm, the ratio of ∠ DAB to ∠ ABC is 1:2, and the intersection point of diagonal AC and BD at point O is to find the length of BD and ac.. It is urgent.)

The diamond ABCD DAB and ABC complement each other, and the ratio of the degree of ∠ DAB and ∠ ABC is 1:2, then ∠ DAB = 60, ∠ ABC = 120. Because the diamond diagonal is vertically bisected, and ∠ DAB and ∠ ABC are bisected, so ∠ ADB = 60, ∠ DAC = 30, the circumference of diamond ABCD is 16cm, so AB = BC = CD = ad = 4, triangle abd is equilateral triangle

In the pyramid p-abcd, the base ABCD is a parallelogram, angle DAB = 60 degrees, ab = 2ad, PD vertical bottom ABCD. (1) it is proved that PA is perpendicular to BD; (2) if P In the pyramid p-abcd, the base ABCD is a parallelogram, angle DAB = 60 degrees, ab = 2ad, PD vertical bottom ABCD. (1) prove PA vertical BD; (2) if PD = ad, find cosine value of dihedral angle a-pb-c

1. Prove ad ⊥ BD by cosine theorem, then BD ⊥ plane pad, get the first question;
2. In plane PBD, if DH ⊥ Pb is made in H, then ah ⊥ edge Pb is formed. When h passes through h, hm / / BC is made in plane PBC and PC is intersected with m, then ∠ AHM is the plane angle of dihedral angle, which is solved in △ AHM

In the parallelogram ABCD, the bisector of angle ABC intersects CD at point E and the bisector of angle ADC intersects AB at point F. this paper tries to prove that the quadrilateral dfbe is a parallelogram

Because in the parallelogram ABCD, angle a = angle c; ad = BC, angle ADF = (1 / 2) angle ADC = (1 / 2) angle ABC = angle EBC
So the triangle ADF is congruent with the triangle CBE, so AF = CE
Because in the parallelogram ABCD, ab | DC, and ab = DC, de = BF
So, de|||||||||||||||||||||||

As shown in the figure, in the parallelogram ABCD, the bisectors of AB = 5, ad = 8, ∠ bad and ∠ ADC intersect BC at e and f respectively, then ef=______ .

∵ AE bisection ∵ bad,
∴∠BAE=∠DAE，
And ∵ ad ∵ CB,
∴∠AEB=∠DAE，
∴∠BAE=∠AEB，
Then be = AB = 5;
Similarly, CF = CD = 5
∴EF=BE+CF-BC=BE+CF-AD=5+5-8=2．
So the answer is: 2

We know: as shown in the figure, point P is a point on the extension line of CD side in the parallelogram ABCD, connecting BP, crossing ad, at point E. explore: when there is a quantitative relationship between PD and CD, △ Abe ≌ △ DPE

When PD = CD, △ Abe ≌ △ DPE
Draw the figure as follows:
It is proved that: ∵ quadrilateral ABCD is a parallelogram
∴AB=CD，AB∥CD，
∴∠BAE=∠PDE，
And ∵ PD = CD,
∴AB=DP，
In △ Abe and △ DPE
∠BAE＝∠PDE
∠AEB＝∠DEP
AB＝DP
≌△ DPE (AAS)

As shown in the figure, in the parallelogram ABCD, the bisector of ∠ ABC intersects ad at point E, bisector of ∠ BCD intersects ad at point F and intersects be at point g. verification: AF = de

Because be, CF are ∠ ABC, ∠ BCD angle bisector, parallel to BC, so ∠ Abe = ∠ AEB, ∠ FCD = ∠ CFD, so Abe = CD = 3 = AE = DF because AE = AF + EF, DF = de + EF, ef

In the triangle ABC, ab = 2Ac, ad bisects the angle BAC, and ad = BD. it is proved that CD is perpendicular to AC

prove:
Take point e of AB and connect De
∵AD=BD
▽ de ⊥ AB, i.e. ≁ AED = 90 ° [isosceles triangle with three lines in one]
∵AB=2AC
∴AE=AC
And ? ead = ∠ CAD [AD bisection ∠ BAC]
AD=AD
∴⊿AED≌⊿ACD（SAS）
∴∠C=∠AED=90º
∴CD⊥AC

In the triangle ABC, angle c = 2, angle B, ad vertical AB, BD = 2Ac

If the right triangle abd is in the circle, BD is the diameter
Make auxiliary line intersection a through midpoint e of BD, and AE is radius
Angle AEC = 2 angle B
Then the angle AEC = angle C
AE=AC
So 2Ac = BD