In the geometry shown in the figure, EA vertical plane ABC, DB vertical plane ABC, AC vertical plane BC and AC = BC = BD = 2ae. M is the midpoint of ab Finding the tangent of the angle between de and plane EMC
① ∵ AE ⊥ plane ABC, BD ⊥ plane ABC, get AE ∥ dB, and plane aedb ⊥ plane ABC
And ∵ AC ⊥ BC and AC = BC, the triangle ABC is isosceles right triangle, and the right angle is ∠ ACB,
If M is the midpoint of AB, then cm ⊥ ab
Aedb and ABC
All the lines on the aedb of the ⊥ cm ⊥ plane
That is cm ⊥ em
②
∵ cm ⊥ aedb
Get cm ⊥ DM
Let AC = BC = BD = 2ae = 2A, EM = √ 3a, MD = √ 6a, ed = 3A
That is ed & sup2; = EM & sup2; + MD & sup2;
The EMD of triangle is right triangle
EM ⊥ MD
And ∵ cm ⊥ DM, the EMC of DM ⊥ plane is obtained, then the angle between de and plane EMC is ∠ DEM,
∴tan∠DEM=DM/EM=√2
As shown in the figure, square ABCD, be, vertical ed, connecting BD and CE, (1) prove that angle EBD = angle ECD? (2) let EB and EC intersect ad at two points F and G, AF = 2fg, explore the quantitative relationship between CG and DG and prove that?
(1) Four point common circle of ∠ bed + ∠ BCD = 180 °, four point common circle of BCDE, i.e. ∠ EBD = ∠ ECD (2) △ AFB ∽ EFD 〉 AF / AB = EF / edbcde, four point common circle of ∠ BEC = ∠ BDC = 45 °, i.e. ∠ BEC = ∠ deceg is the bisector of ∠ fed angle, i.e. EF / ed = FG / DG, i.e. AF / AB = FG / dgab = DG = 2, i.e. CD / DG = 2 〈 CG / DG = sqrt
In the triangle ABC, angle B plus angle c is equal to twice the angle a, CF is vertical AB, be is vertical AC, D is the midpoint of BC, and the triangle EFD is a triangle. It is proved that
From ∠ B + ∠ C = 2 ∠ a, it is concluded that ∠ B + ∠ C + ∠ a = 3 ∠ a = 180 °, a = 60 °. In △ BCF, ∠ BFC = 90 °, DF = 1 / 2 · BC, △ BEC, ∠ BEC = 90 °, de = 1 / 2 · BC, de = DF. ∵ BD = FD, ∵ BDF = 180 ° - 2 ∠ B, ∵ - ed = CD, ∵ CDE =
BC = DC, CA bisector angle BCD, prove: (1) vertical BD; (2) AB = ad
Because CA is bisector
So angle BCA = angle DCA
Because in triangle ABC and triangle DCA
So CB = CD
Angle BCA = angle DCA
CA=CA
So,
1 because angle ADC = angle CBA
Angle ACB = angle ACD
So CA vertical BD
In the triangular pyramid a-bcd, e is the midpoint of BC, ab = ad, BD vertical DC, to prove AE vertical BD
Let AB = 2A, BC = 2A, BD = a, AC = 2 * radical 2a, DC = radical 3a. In the triangle ADC, find AF = (4 * radical 5) / 5 in the triangle abd
As shown in the figure, ab = DC, AC = dB, which pairs of triangles are congruent in the figure? Why?
There are four pairs of congruent triangles
Delta ade and delta ADF, delta ADB and delta ADC
△ Abe and △ ACF △ ABF and △ ace
Enough details!
As shown in the figure, in the known isosceles trapezoid ABCD, ad ‖ BC, ab = DC, AC and BD intersect at point O. please find a pair of congruent triangles in the figure and prove them
The △ ABC ≌ DCB (2 points) proves that: in the isosceles trapezoid ABCD, ad ‖ BC, ab = DC ≌ ABC = DCB (4 points) in △ ABC and △ DCB, ab = DC ≌ ABC = dcbbc = BC ≌ ABC ≌ DCB (7 points) (Note: the answer is not unique)
As shown in the figure, in the trapezoidal ABCD, ab ∥ CD, ad ⊥ AC, ad = AC, DB = DC, AC, BD intersect at point e to find the degree of ∠ BDC
Make AE ⊥ DC at point E, BF ⊥ DC at point F
Then AE ‖ BF
The quadrilateral aefb is a rectangle
∴AE=BF
∵∠DAC=90°,AD=AC
∴AE=1/2DC
∵DC =DB
∴BF=AE=1/2BD
∴∠BDC=30°
As shown in the figure, in the trapezoidal ABCD, ab ∥ CD, AC ⊥ BD, AC = ad = √ 2, DB = DC, AC and BD intersect at point E, AB is obtained
RT
Let am ⊥ CD be m, and BN ⊥ CD be n. obviously, the quadrilateral amnb is a rectangle, BN = am, ab = Mn
Because AC ⊥ BD, AC = ad = √ 2, so CD = 2, am = DM = CD / 2 = 1
Because BD = CD = 2, BN = am = 1, BN ⊥ CD, so DN = √ 3,
So, Mn = dn-dm = √ 3-1, so AB = √ 3-1
As shown in the figure, it is known that ∠ a = ∠ d = 90 ° AC = BD, and it is proved that: (1) AB = DC; (2) ob = OC
It is proved that: (1) in RT △ ABC and RT △ DCB, BC = CBAC = BD ≌ RT △ ABC ≌ RT △ DCB (HL), ≌ AB = DC; (2) in RT △ ABC and RT △ DCB, BC = CBAC = BD ≌ RT △ ABC ≌ RT △ DCB, ≌ ACB = DBC, ≌ ob = OC