Same as the linked graph

It is proved that because AB is vertical to AC and ad is vertical to AE, so ∠ BAC = ∠ DAE = 90 ° so ∠ BAC + ∠ CAE = ∠ DAE + ∠ CAE, that is ∠ BAE = ∠ CAD in △ BAE and △ CAD, ab = AC ∠ BAE = ∠ cadae = ad, so △ BAE ≌△ CAD (SAS), so ∠ B = ∠ C because AB is vertical AC, so ∠ B + ∠ AGB = 90 ° so ∠ C +

In △ ABC, if Tana = 1 3, C = 150 ° and BC = 1, then AB = () A. Ten B. 2 Ten Ten C. 2 Ten D. Ten Two

∵ in △ ABC, Tana = 1
Three
∴sinA=1
10=
Ten
10，
Sine can be obtained according to the theorem
BC
sinA＝AB
sinC，
∴1
Ten
10＝AB
One
Two
∴AB=
Ten
2，
Therefore, D

In the right triangle ABC, the angle ACB = 90 degrees, CD is perpendicular to AB and D, De is perpendicular to AC and e. it is proved that CE is more than AE = BC

The question is not complete. It should be CE / AE = BC ^ 2 / AC ^ 2. The certificate above is also wrong,
De ⊥ AC, so de / / BC,
According to the proportional property of parallel line segments,
CE/AE=BD/AD,（1）
Obviously, △ CDB ∽ AED,
BD/ED=BC/AD,
BD=BC*ED/AD,
Substitute (1),
CE/AE=（BC/AD）*ED/AD,（2）
△AED∽△ACB,
ED/BC=AD/AB,
Ed / ad = BC / AB, (more than),
Substitute (2),
CE/AE=（BC/AD）*（BC/AB）=BC^2/(AD*AB),
△ADC∽△ACB,
AC/AB=AD/AC,
AC^2=AD*AB,
∴CE/AE=BC^2/AC^2.

As shown in the figure, in RT △ ABC, ∠ ACB = 90 °, AC = 4, BC = 3, D is a point on the hypotenuse AB, and the parallelogram cdeb is made with CD and CB as the sides=______ The parallelogram cdeb is rhombic

As shown in the figure, connect CE to ab at point o
∵ RT △ ABC, ∵ ACB = 90 °, AC = 4, BC = 3,
∴AB=
Ac2 + BC2 = 5 (Pythagorean theorem)
If the parallelogram cdeb is rhombic, CE ⊥ BD, OD = ob, CD = CB
∵1
2AB•OC=1
2AC•BC，
∴OC=12
5．
In RT △ BOC, according to Pythagorean theorem, ob is obtained=
BC2−OC2=
32−(12
5)2=9
5，
∴AD=AB-2OB=7
5．
So the answer is: 7
5．

As shown in the figure, in the right triangle ABC, the angle ACB = 90 degrees, CD is perpendicular to D, let AC = B, BC = a, ab = C, CD = h, try to explain the composition of a + B, h, C + H three sides A triangle is a right triangle

S△ABC=1/2ab=1/2ch
∴ab=ch
(a+b)²=a²+2ab+b²
∵a²+b²=c² ab=ch
Therefore, the original formula = C 2 + 2CH
(c+h)²=c²+2ch+h²
∴(a+b)²+h²=(c+h)²
So the triangle composed of a + B, h, C + H is a right triangle

As shown in the figure, s is a point outside the plane where the regular triangle ABC with side length a is located, SA = sb = SC = a, e and F are the midpoint of AB and SC, then the angle formed by SA and EF of the straight line SA and EF is______ .

Take the midpoint o of AC, connect EO and fo, take the midpoint P of BC, connect SP and AP,
∵ s is a point outside the plane ABC of the equilateral triangle, and SA = sb = SC = AB = a,
∴SP⊥BC，AP⊥BC，
ν BC ⊥ plane ASP,
∴BC⊥AS．
∵ E and F are the midpoint of SC and ab respectively,
Therefore, of and OE are median lines, so OE ‖ SA, OE ‖ BC, and OE = 1
2SA=1
2a，OE=1
2BC=1
2a，
⊥ fo, and EO = fo, ﹤ FeO is the angle formed by the straight line EF and SA,
∴∠FEO=45°．
So the answer is: 45 degrees

As shown in the figure, s is a point outside the plane where the regular triangle ABC with side length a is located, SA = sb = SC = a, e and F are the midpoint of AB and SC, then the angle formed by SA and EF of the straight line SA and EF is______ .

Take the midpoint o of AC, connect EO, fo, take the midpoint P of BC, connect SP, AP, ∵ s is the point outside the plane of the equilateral triangle, and SA = sb = SC = AB = a, ⊥ sp ⊥ BC, AP ⊥ BC, ᚉ BC ⊥ plane ASP, ∵ BC ⊥ as. ∵ E and F are the midpoint of SC and ab respectively, ∵ so of and OE are respectively the median lines, so OE ∥

As shown in the figure, s is a point outside the plane where the regular triangle ABC with side length a is located, SA = sb = SC = a, e and F are the midpoint of AB and SC, then the angle formed by SA and EF of the straight line SA and EF is______ .

As shown in the figure, SA ⊥ sb, sb ⊥ SC, SA ⊥ SC, and the angles formed by SA, Sb, SC and bottom ABC are A1, A2 and A3 respectively, and the areas of ⊥ SBC, △ sac and △ SAB of three sides are S1, S2 and S3 respectively. A conjecture of space case is given by analogy with the sine theorem in triangle

The solution is in △ def,
Get D
sinD=e
sinE=f
sinF．
So, analogy with the sine theorem in triangles,
In tetrahedral s-abc,
We guess S1
sinα1=S2
sinα2=S3
Sin α 3 is established

As shown in the figure, point s is outside the plane ABC, sb ⊥ AC, sb = AC = 2, e and F are the midpoint of SC and ab respectively, then the length of EF is () A. 1 B. Two C. Two Two D. 1 Two

Take the midpoint D of BC to connect ed and FD
∵ E and F are the midpoint of SC and ab respectively, and point D is the midpoint of BC
∴ED∥SB，FD∥AC
If sb ⊥ AC, sb = AC = 2, then EDF is isosceles right triangle
Then ed = FD = 1, that is EF=
Two
Therefore: B

Same as the linked graph