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

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