As shown in the figure, △ ABC, point D is on AC, and ab = ad, ∠ ABC = ∠ C + 30 °, then ∠ CBD=______ Degree

∵AB=AD
∴∠ABD=∠ADB
∵∠ADB=∠C+∠CBD
∴∠ABD=∠C+∠CBD
∴∠ABC=∠ABD+∠CBD=2∠CBD+∠C
It is known that ∠ ABC = ∠ C + 30 °
∴2∠CBD+∠C=∠C+30°
That is ∠ CBD = 15 °
Therefore, fill in 15

As shown in the figure, ⊙ o radius is 2, chord BD = 2 3, a is the midpoint of arc BD, e is the midpoint of chord AC, and on BD, the area of quadrilateral ABCD is calculated

Connect OA to BD at point F, connect ob,
∵ OA is on the diameter and point a is the midpoint of arc BD,
∴OA⊥BD，BF=DF=
Three
In RT △ BOF
From Pythagorean theorem, of2 = ob2-bf2 is obtained
OF=
22−(
3)2=1
∵OA=2
∴AF=1
∴S△ABD=2
3×1
2=
Three
∵ point E is the midpoint of AC
∴AE=CE
And ∵ ∵ ade and ∵ CDE are the same
∴S△CDE=S△ADE
∵AE=EC，
∴S△CBE=S△ABE．
∴S△BCD=S△CDE+S△CBE=S△ADE+S△ABE=S△ABD=
Three
﹤ s quadrilateral ABCD = s △ abd + s △ BCD = 2
3．

As shown in the figure, ⊙ o radius is 2, chord BD = 2 3, a is the midpoint of arc BD, e is the midpoint of chord AC, and on BD, the area of quadrilateral ABCD is calculated

As shown in the figure, the radius of circle O is 2, chord BD = 2, root sign 3, a is the midpoint of arc BD, e is the midpoint of chord AC, and on BD, find the face of quadrilateral ABCD You also answered, but I want to know why AF = of, how to prove it

Where's the picture, man

As shown in the figure, ⊙ o radius is 2, chord BD = 2 3, a is the midpoint of arc BD, e is the midpoint of chord AC, and on BD, the area of quadrilateral ABCD is calculated

Given the point ABCD on the circle O, a is the midpoint of BD arc, AC BD intersects with point E, and E is the midpoint of AC chord, the radius of circle is 2, BD is 2 times the root sign 2. Find the quadrilateral ABCD surface

Make a tangent through a, and make a straight line parallel to the tangent line through C
It is found that the two lines of a C and BD are a group of parallel lines
If AE = AC, then the distances between the three parallel lines, that is, the triangle abd and the triangle CBD are all BD, then the heights are equal
If Ao intersects BD with F, then of = radical 2, AF = 2-radical 2
So the quadrilateral area is BD times AF

As shown in the figure, it is known that in ▱ ABCD, the parallel line Mn of AC intersects Da, the extension line of DC intersects with m, N, crosses AB, BC at P, Q, and proves that QM = NP

It is proved that: ∵ quadrilateral ABCD is a parallelogram
∴MD∥BC，AB∥ND，
∵MN∥AC，
∴MQ∥AC，AM∥QC，PN∥AC，AP∥CN，
The quadrilateral amqc and quadrilateral APNC are parallelograms,
∴MQ=AC，PN=AC，
∴QM=NP．

As shown in the figure, in the parallelogram ABCD, the parallel line Mn of AC crosses the extension line of DA at m, the extension line of crossing DC at n, and the intersection of AB and BC in P, Q 1) Please indicate the number of parallelogram in the figure and explain the reason (2) Can MP and QN be equal?

1. Three parallelograms:
ABCD、 AMQC、 APNC
2、MP=QN
Conclusion: from the parallelogram amqc, MQ = AC
From the parallelogram APNC to know PN = AC
∴MQ=PN
That is MP + PQ = PQ + QN
∴MP=QN

AB is the diameter of circle O, AC is the chord, angle a = 30 degrees, D is on the extension line of AB, DC = AC

Connecting OC, because Ao = Co, angle a = angle ACO, and angle a = 30, so angle ACO = 30 degrees, so angle cod = 60 degrees, and angle d = 30 degrees, so angle DCO = 90 degrees, so OC is perpendicular to CD, that is, DC is tangent line of circular O

As shown in the figure, the diameter ab of ⊙ o is 4; the straight line Mn passing through point B is the tangent line of ⊙ o; D and C are two points on ⊙ o, connecting ad, BD, CD and BC (1) It was proved that ∠ CBN = ∠ CDB; (2) If DC is a bisector of ∠ ADB and ∠ DAB = 15 °, find the length of DC

As shown in the figure, △ ABC, point D is on AC, and ab = ad, ∠ ABC = ∠ C + 30 °, then ∠ CBD=______ Degree