As shown in the figure, in the parallelogram ABCD, AE and CF bisect ∠ bad and ∠ DCB respectively, and intersect BC and ad at point E and point F It is shown that (1) △ Abe is isosceles triangle; （2） The quadrilateral AECF is a parallelogram

As shown in the figure, in the parallelogram ABCD, AE and CF bisect ∠ bad and ∠ DCB respectively, and intersect BC and ad at point E and point F It is shown that (1) △ Abe is isosceles triangle; （2） The quadrilateral AECF is a parallelogram

It is proved that: (1) ∵ quadrilateral ABCD is a parallelogram,
∴∠BAD=∠DCB，AD∥BC，
∵ AE and CF are equally divided into ∵ bad and ∠ DCB respectively,
∴∠BAE=∠DAE=1
2∠BAD，
∴∠DAE=∠AEB，
∴∠BAE=∠AEB，
∴BA=BE，
The △ Abe is an isosceles triangle;
(2) In the same way, it can be proved that △ DCF is an isosceles triangle,
∴DF=DC，
According to (1), Ba = be,
∵AB=CD，AD=BC，
∴DF=BE，
∴AF=EC，
∵AF∥EC，
The quadrilateral aecf is a parallelogram

In the triangle ABC, the angle ABC = 45 degrees, ad is the bisector of angle BAC, EF is the vertical bisector of AD, the extension line of intersection BC is at point F, and the degree of angle CAF is --------?

Because EF is the vertical bisector of AD, so △ ADF is an isosceles triangle, AF = DF, ∠ ADF = ∠ DAF
That is ∠ abd + ∠ bad = ∠ DAC + ∠ CAF, and ∠ bad = ∠ DAC, so ∠ CAF = ∠ bad = 45 °

It is known that in △ ABC, ∠ BAC = 90 °, ad ⊥ BC at point D, be bisection ∠ ABC, intersection ad at point m, an bisection ∠ DAC, intersection BC at point n Verification: the quadrilateral amne is rhombic

Proof: ∵ ad ⊥ BC,
∴∠BDA=90°，
∵∠BAC=90°，
∴∠ABC+∠C=90°，∠ABC+∠BAD=90°，
∴∠BAD=∠C，
∵ an bisection ∠ DAC,
∴∠CAN=∠DAN，
∵∠BAN=∠BAD+∠DAN，∠BNA=∠C+∠CAN，
∴∠BAN=∠BNA，
∵ be bisection ∵ ABC,
∴BE⊥AN，OA=ON，
Similarly, OM = OE,
The quadrilateral amne is a parallelogram,
The parallelogram amne is a diamond

It is known that: ab = AC, BD ⊥ AC in the figure △ ABC, and it is proved that ∠ CBD = half ∠ a Just an hour

Let the midpoint of BC be e
BE=EC AE=AE AB=AC
△ABE≌△ACE
∠AEB=∠AEC=90°
∠EAC+∠C=90°=∠CBD+∠C
So ∠ EAC = ∠ CBD = ∠ EAB = 1 / 2 ∠ a
I wish you every day to learn, come on!

In the triangle ABC, given the angle a = 30 degrees and the angle CBD = 90 degrees, calculate the number of angles of angle BCE?

Where are the pictures

In the triangle ABC, the angle a = 35 degrees, the angle CBD = 115 degrees

180°-35°-115°=30°
180°-30°=150°

In the triangle ABC, we know ∠ a = 30 ° and ∠ CBD = 90 ° to find the degree of ∠ BCE

∠ABC=180-∠CBD=180°90°=90°，
∠ACB=180°-∠A-∠ABC，
=180°-30°-90°，
=60°，
∠BCE=180°-∠ACB，
=180°-60°，
=120°．
A: BCE is 120 degrees

Triangle ABC, AD is perpendicular to BC, angle B is equal to angle C of 2 times, and it is proved that CD is equal to AB+BD with the knowledge of axisymmetric figure

It is proved that the axisymmetric △ AED of △ abd is made by taking ad as the symmetry axis
That is ∠ B = ∠ AED = 2 ∠ C, BD = De
In △ ace, ∠ AED = ∠ C + ∠ CAE
That is, ∠ C = ∠ CAE, △ AEC is isosceles triangle
Therefore, AE = CE = ab
That is CD = de + CE = AB + BD
The proof is complete

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．

[known online; as shown in the figure, in the parallelogram ABCD, Mn ∥ AC, respectively intersects the extension lines of Da and DC at point m, N, intersection AB, CB at point P, Q to verify; MQ = n

prove:
∵AD‖BC
∴∠MAP=∠B
∵AB‖CD
∴∠NCQ=∠B
∴∠NCQ=∠MAP
∵MN‖AC
The quadrilateral amqc and quadrilateral APNC are parallelograms
∴AM=CQ,AP=CN
∴△AMP≌△CNQ
∴MP=NQ
∴MQ=NP