As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 108 °, D is on AC and BC = AB + CD, the proof is: BD bisects ∠ ABC

It is proved that: take a point E on BC, let be = AB, then CE = CD, because AB = AC, angle BAC = 108 ', then angle ABC = angle ACB = 36 degrees. (this degree is very important, which is the golden section ratio of triangle). Because CE = CD, angle c (ACB) = 36 degrees, so: angle CDE = angle CED = 72 degrees, because AB = be, angle Abe = 36 degrees, so angle BAE = angle B

It is known that ab = AC, ∠ a = 108 ° as shown in the figure, BD bisects ∠ ABC and intersects AC with D. It is proved that BC = AB + CD

In △ abd and △ EBD, be = BA ∠ abd ＝ ebdbd = BD (common edge), △abd ≌ △ EBD. (SAS) ≠ bed = ∠ a = 108 °, ADB = ∠ EDB. And ≌ AB = AC, ∠ a = 108 · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·

It is known that ab = AC, ∠ a = 108 ° as shown in the figure, BD bisects ∠ ABC and intersects AC with D. It is proved that BC = AB + CD

In △ abd and △ EBD, be = Ba, be = Ba, be = Ba, be = Ba, be = Ba, be = Ba, be = Ba, be = Ba, connect de \\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\it's not easy °-18°-108°=54°．∴∠CDE=180°-∠ADB-∠EDB=180°-54°-54°=72°．∴∠DEC=180°-∠DEB=180°-108°=72°．∴∠CDE=∠DEC．∴CD=CE．∴BC=BE+EC=AB+CD．

It is known that ab = AC, ∠ a = 108 ° as shown in the figure, BD bisects ∠ ABC and intersects AC with D. It is proved that BC = AB + CD

In the triangle ABC, ab = AC, angle a = 108 degrees, BD bisector angle ABC intersects AC with D, proving BC = AB + CD

In space quadrilateral ABCD, ab = BC = CD = Da = A. diagonal AC = a, BD = root 2 · a, find the size of a-bd-c, find the cosine value of plane angle of b-ac-d

As shown in the figure, ⊿ abd & nbsp; ⊿ CBD are isosceles right triangles, O is the midpoint of BD, AOC is the plane angle of
Ao = co = A / √ 2. & nbsp; & nbsp; AC = A & nbsp; & nbsp; & nbsp; & nbsp; AOC = 90 & amp; ordm; & nbsp; & nbsp; a-bd-c is a dihedral angle
Let p be the midpoint of AC (not shown in the figure) & nbsp; ∠ BPD be the plane angle of dihedral angle b-ac-d
BP = DP = √ 3A / 2. BD = √ 2a, & nbsp; & nbsp; & nbsp; from cosine theorem & nbsp; & nbsp; Cos ∠ BPD = - 1 / 3

In the quadrilateral ABCD, ab = BC = CD = Da = BD = 1, the value range of AC is______

In the quadrilateral ABCD, it becomes a space tetrahedron. When a is close to the coincidence of C, the distance of AC is close to 0; when ABCD is close to a plane figure, the maximum distance of AC is close to 3, so AC ∈ (0,3), so the answer is: (0,3)

In the space quadrilateral ABCD, ab = BC = CD = Da = 1, if BD = √ 2, then the value range of AC is ()

AB=BC=CD=DA=1,BD=√2,
Both △ abd and △ CBD are right angle isosceles triangles
0 ＜ AC ≤ √ 2 (when four points of ABCD are in the same plane, AC is the maximum √ 2)

In space quadrilateral ABCD, if AB = BC = CD = Da, find the common vertical segment of AC and BD

Let e be the midpoint of BD and f be the midpoint of AC
⊿ abd isosceles, ⊥ AE ⊥ BD
⊿ CBD isosceles, ⊥ CE ⊥ BD
⊥ BD ⊥ plane AEC, BD ⊥ EF
∵⊿ABD≌⊿CBD.（S,S,S）.
AE = CE. ⊿ EAC isosceles
∴EF⊥AC.
EF is the common vertical of BD and AC

In the tetrahedral ABCD, if AC and BD form an angle of 60 ° and AC = BD = a, then the area of the quadrilateral connecting the midpoint of AB, BC, CD and DA is___ ．

Take the midpoint of AB, BC, CD, DA as e, F, G, h respectively, and connect eh, EF, FG, Hg, so we get: EH is the median line of △ abd, so eh ‖ BD, and eh = 12bd. Similarly, FG ‖ BD, and FG = 12bd,. So eh ‖ FG, and eh = FG. So efgh is a parallelogram

As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 108 °, D is on AC and BC = AB + CD, the proof is: BD bisects ∠ ABC