As shown in the figure, the vertex angle of the isosceles △ ABC is 50 ° and ab = AC. taking AB as the diameter, make a semicircle and intersect BC at point D and AC at point E BD、 De and The degree of the center angle to which AE is directed

As shown in the figure, the vertex angle of the isosceles △ ABC is 50 ° and ab = AC. taking AB as the diameter, make a semicircle and intersect BC at point D and AC at point E BD、 De and The degree of the center angle to which AE is directed

Connect be, ad,
∵ AB is the diameter of the circle,
∴∠ADB=∠AEB=90°，
∴∠ABE=90°-50°=40°，
AD⊥BC，
∵AB=AC，∠BAC=50°，
∴∠BAD=∠DAC=1
2∠BAC=25°，
According to the circular angle theorem, the degree of the center angle of arc BD is 2 ∠ DAB = 50 ° and that of Arc De is 2 ∠ DAE = 50 ° and that of arc AE is 2 ∠ BAE = 80 °

As shown in the figure, ∠ a = 60 ° in △ ABC, make ⊙ o with BC as the diameter, cross AB, AC with D and e respectively, (1) Results: ab = 2ae; (2) If AE = 2, CE = 1, find BC

(1) Proof: connect be,
∵ BC is the diameter of ⊙ o,
∴∠BEC=90°，
That is ∠ AEB = 90 °,
∵∠A=60°，
∴∠ABE=30°，∴AB=2AE；
（2）∵AE=2，
∴AB=2AE=4，
∴BE=
AB2−AE2=2
3，
∵CE=1，
∴BC=
BE2+CE2=
13．

The three vertices a, B and C of triangle ABC are all on circle O, and E is the midpoint of arc BC. It is proved that ab * be = AE * BD

[D is the intersection of AE and BC]
prove:
∵ e is the midpoint of arc BC
ν arc be = arc CE
Ψ BAE = ∠ CAE
∵ EBC = ∵ CAE
∴∠BAE=∠EBC
And ∵ bea = ∵ DEB [common angle]
∴⊿BAE∽⊿DBE（AA‘）
∴AB/BD=AE/BE
It was transformed into AB × be = AE × BD

In the triangle ABC, if the angle c = 90 degrees, CA = CB, ad bisect angle cab, intersection BC to D, de perpendicular to AB and ab = 6, what is the perimeter of triangle DEB? Deb is a triangle

By using the AAS of triangle open total, we can judge that ACD of triangle is equal to ade of triangle
AE=AC=BC CD=DE
The perimeter of the triangle DEB is
DE+DB+BE
=CD+（BC-CD）+BE
=BC+BE
=AE+BE
=6

As shown in the figure, in △ ABC, ∠ C = 90 °, ad bisects ∠ cab, intersects CB at point D, and passes through point D as de ⊥ AB at point E (1) Verification: △ ACD ≌ △ AED; (2) If ∠ B = 30 ° and CD = 1, find the length of BD

(1) It is proved that: ∵ ad bisection ∵ cab, de ⊥ AB, ∵ C = 90 °,
∴CD=ED，∠DEA=∠C=90°，
∵ in RT △ ACD and RT △ AED
AD＝AD
CD＝DE
∴Rt△ACD≌Rt△AED（HL）；
（2）∵DC=DE=1，DE⊥AB，
∴∠DEB=90°，
∵∠B=30°，
∴BD=2DE=2．

It is known that: as shown in the figure, in the triangle ABC, the angle a = the angle ABC, and the straight line EF intersect the extension of the edges AB, AC and CB of the triangle ABC respectively The line is at point D, e and F, and the angle F + angle FEC = 2 angle A

The title seems to be wrong
Angle ECF = angle a + angle B = 2 angle a
Angle F + angle FEC + angle ECF = 180 degrees

As shown in the figure, in right angle trapezoid ABCD, ∠ ABC = 90 °, ad ‖ BC, ab = BC, e is the midpoint of AB, CE ⊥ BD (1) Results: be = ad; (2) Verification: AC is the vertical bisector of ED; (3) Is △ DBC an isosceles triangle? And explain the reason

(1) It is proved that ∵ ABC = 90 ° BD ⊥ EC,
∴∠1+∠3=90°，∠2+∠3=90°，
∴∠1=∠2，
In △ bad and △ CBE,
∠2＝∠1
BA＝CB
∠BAD＝∠CBE＝90° ，
∴△BAD≌△CBE（ASA），
∴AD=BE．
(2) It is proved that ∵ e is the midpoint of ab,
∴EB=EA，
∵AD=BE，
∴AE=AD，
∵AD∥BC，
∴∠7=∠ACB=45°，
∵∠6=45°，
∴∠6=∠7，
And ∵ ad = AE,
⊥ De, and EM = DM,
AC is the vertical bisector of ED;
(3) Δ DBC is an isosceles triangle (CD = BD)
The reasons are as follows.
∵ from (2), CD = CE, from (1), CE = BD,
∴CD=BD．
The △ DBC is an isosceles triangle

As shown in the figure, in right angle trapezoid ABCD, ∠ ABC = 90 °, ad ‖ BC, ab = BC, e is the midpoint of AB, CE ⊥ BD (1) Results: be = ad; (2) Verification: AC is the vertical bisector of ED; (3) Is △ DBC an isosceles triangle? And explain the reason

(1) It is proved that ∵ ABC = 90 ° BD ⊥ EC,
∴∠1+∠3=90°，∠2+∠3=90°，
∴∠1=∠2，
In △ bad and △ CBE,
∠2＝∠1
BA＝CB
∠BAD＝∠CBE＝90° ，
∴△BAD≌△CBE（ASA），
∴AD=BE．
(2) It is proved that ∵ e is the midpoint of ab,
∴EB=EA，
∵AD=BE，
∴AE=AD，
∵AD∥BC，
∴∠7=∠ACB=45°，
∵∠6=45°，
∴∠6=∠7，
And ∵ ad = AE,
⊥ De, and EM = DM,
AC is the vertical bisector of ED;
(3) Δ DBC is an isosceles triangle (CD = BD)
The reasons are as follows.
∵ from (2), CD = CE, from (1), CE = BD,
∴CD=BD．
The △ DBC is an isosceles triangle

As shown in the figure, in the isosceles trapezoid ABCD, ad is parallel to BC, angle c is equal to 60 degrees, BD bisects angle ABC, and it is proved that ad = 1 / 2BC

It is proved that: ? quadrilateral ABCD is trapezoid; ᙽ AD / / BC; (two bases of isosceles trapezoid are parallel)  ADC + ∠ C = 180 °; (two straight lines are parallel, and the inner angles of the same side are complementary) ? C = 60; ? ADC = 120 °; ∵ C = 60 ° and trapezoid ABCD is isosceles trapezoid; ᚅ AB = CD; (the two waists of isosceles trapezoid are equal) ?

As shown in the figure, the quadrilateral ABCD is a right angle trapezoid, ∠ ABC = ∠ bad = 90 ° SA ⊥ plane ABCD, SA = AB = BC = 1, ad = 1 2． (1) Find the cosine of angle formed by SC and ASD; (2) Find the cosine of the angle formed by plane SAB and plane SCD

(1) Make CE ∥ AB to the extension line of ad at E,
∵AB⊥AD，
∴CE⊥AD．
And ∵ SA ⊥ surface ABCD,
∴CE⊥SA，SA∩AD=A，
⊥ face sad, se is the projection of SC in sad,
Ψ CSE = θ is the angle between SC and ASD,
Easy to get se=
2，SC=
3，
In RT △ CES, cos θ = CE
SC=
Six
Three
(2) From SA ⊥ face ABCD, know face ABCD ⊥ face SAB,
The projection of △ SCD on the surface SAB is △ SAB,
The area of △ SAB is S1 = 1
2×SA×AB=1
2，
Let the midpoint of SC be m, ∵ SD = CD=
Five
2，
∴DM⊥SC，DM=
Two
Two
The area of △ SCD is S2 = 1
2×SC×DM
Six
Four
Let the angle between plane SAB and plane SCD be φ,
From the area projective theorem, cos φ = s △ SAB is obtained
S△SCD=
Six
Three