In the triangle ABC, ad is the middle line of BC side, P is the middle point of AD, extending BP to meet AC at point F, indicating Pb = 3pf

In the triangle ABC, ad is the middle line of BC side, P is the middle point of AD, extending BP to meet AC at point F, indicating Pb = 3pf

Passing through point D, make de parallel to BP and intersect AC at point E
Because P is the midpoint of AD
So PF: de = AP: ad = 1:2, PF = 1 / 2 De
And D is the midpoint of BC
So de: BF = CD: BC = 1:2
So BF = 2DE = 4pf
So Pb = 3pf

As shown in the figure, in rectangular ABCD, fold △ ABC around AC to △ AEC position, CE and ad intersect at point F, as shown in the figure. Explain EF = DF

It is proved that AE = AB = DC according to the properties of rectangle, according to the equality of opposite vertex angles, ∠ EFA = ∠ DFC, and ∠ AEC = ∠ ADC = 90 °. From AAS, △ AEF ≌ △ CDF {EF = DF

In rectangular ABCD, ab = 4, BC = 8, take diagonal AC as symmetry axis, fold △ ABC along AC, point B falls at point E, CE and ad intersect at F. calculate the circumference of △ AFC
To specific process. No map.. we make do with painting it, not a very complex map

From the problem, we can see that △ ABC ≌ △ AEC, AC bisection ∠ BCE, and because ad is parallel to BC, we can get △ AFC as isosceles triangle, so AF = CF, let AF be x, then DF is 8-x, and CF is also X. in △ FDC, using Pythagorean theorem, we can know that x = 5, that is, AF = FC = 5, because AC is diagonal, it is easy to know that AC = quadruple root 5, So the perimeter of △ AFC is 10 + quadruple root 5

If the isosceles △ ABC vertex angle ∠ BAC = 120 ° and the waist length is 10, what is the area of the triangle

From a to ad ⊥ BC,

If the waist length of isosceles △ ABC is ab = 2, the vertex angle ∠ BAC = 120 ° and the square area with BC as the side is ()
A. 3B. 12C. 274D. 163

Let ad ⊥ BC be d. ⊥ AB = AC, ∠ BAC = 120 ° and ⊥ abd = 30 ° and ⊥ ad = 12ab = 1. According to the Pythagorean theorem, we get BD = 3. According to the three lines of isosceles triangle, we get BC = 2bd = 23, then the area of the square with BC side length is (23) 2 = 12, so we choose B

If P is a point outside the plane of equilateral triangle ABC, PA = Pb = PC = 23, and the side length of △ ABC is 1, then the angle between PC and plane ABC is ()
A. 30°B. 45°C. 60°D. 90°

Take the midpoint D of AB, connect PD and CD, ∵ PA = Pb, D is the midpoint of AB, ≁ PD ⊥ AB, similarly, we can get that CD ⊥ ab ⊥ PD and CD are intersecting lines in plane PCD ≁ ab ⊂ plane ABC, ≁ plane PCD ⊥ plane ABC, so we can get that the projection of line PC in plane ABC is straight line CD, ∽ PCD is straight line PC

In a triangular pyramid p-abc, the three sides PA, Pb and PC are perpendicular to each other, and H is the perpendicularity of △ ABC

It is proved that: (1) connecting ah and extending BC to a point E, connecting pH, because PA, Pb and PC are perpendicular to each other, PA ⊥ plane PBC can be obtained, and BC ⊂ plane PBC, ⊥ BC ⊥ PA and H are the perpendicular of triangle ABC, so AE ⊥ BC, AE ∩ PA = a, ⊥ BC ⊥ plane PAE, and pH ⊂ plane PAE, ⊥ pH ⊥ BC can be proved

P is a point outside the plane ABC, PA vertical Pb Pb vertical PC vertical PA pH vertical plane ABC, h is the perpendicular center of triangle ABC, and the perpendicular center is the intersection point of high

It is proved that: connect BH, extend BH to AC to e, connect ah, extend ah to BC to F
∵PB⊥PA,PB⊥PC
⊥ Pb ⊥ PAC
∴PB⊥AC
∵ pH ⊥ surface ABC
∴PH⊥AC
⊥ AC ⊥ PBE
∴AC⊥BE
Similarly, AF ⊥ BC can be proved
The point h is the perpendicular of the triangle ABC

P is a point outside the plane of triangle ABC, PA, Pb and PC are perpendicular to each other, and pH is perpendicular to H. it is proved that 1 / pa2 + 1 / Pb2 + 1 / PC2 = 1 / pH2

Then we know pH ⊥ ch, and PC ⊥ PD, CD and PD ⊥ ab. then we have pH ^ 2 / PC ^ 2 = sin ^ 2 (angle PCD) = PD ^ 2 / CD ^ 2 = HD * CD / CD ^ 2 = HD / CD = [HD * AB / 2] / [CD * AB / 2] = s △ ha

As shown in the figure, if the plane ABC ⊥ plane abd, ∠ ACB = 90 °, CA = CB, △ abd is an equilateral triangle, what is the tangent value of the plane angle of the dihedral angle c-bd-a

Take the midpoint o of AB, connect Co, make oh ⊥ BD, connect ch ⊥ CA = CB, ∩ plane ABC ⊥ plane abd = AB, ∩ Co ⊥ plane abd, ≁ oh ⊥ BD ⊥ ch ⊥ BD ∩ Cho is the plane angle of dihedral angle c-bd-a. let CA = 2A, then ∩ ACB = 90 °, CA = CB, ∩ co = 2A ⊥ ab