In the diamond ABCD with side length of 2, ∠ bad = 60 °, e is the midpoint of AB, P is the upper moving point of AC, and the minimum value of Pb + PE is obtained

In the diamond ABCD with side length of 2, ∠ bad = 60 °, e is the midpoint of AB, P is the upper moving point of AC, and the minimum value of Pb + PE is obtained

Solution] connects de and AC at point P, then point P is the solution
That is: the value of Pb + PE is the smallest
[proof] take any point P 'on AC and connect p'e and p'b
In triangle dp'e: p'b + p'e > de
Because the quadrilateral ABCD is a diamond, the triangle abd is an isosceles triangle, and the angle bad is 60 degrees
So: the triangle abd is an equilateral triangle
The diamond is symmetrical according to the AC axis, so PD = Pb
So: de = Pb + PE
There is the above proof: for any point P ', there is p' B + p 'E > de, that is, for P point on AC, De is the shortest
So: the minimum value of Pb + PE is De, the length is: 2sin60 ° = root 3

In diamond ABCD, if the angle ADC = 120 degrees, then BD: AC=

Because: angle ADC = 120 degrees, so the angle DAB = 60
So: Triangle ADB is equilateral triangle, that is: BD = ad
Let the intersection of BD and AC be o, then OA = root 3 / 2 * ad, (which can be obtained according to Pythagorean theorem.)
So: AC = 2oa = root 3aD
BD: AC = ad: root 3aD = 1: root 3

In diamond shaped ABCD, ∠ ADC = 120 °, then BD is equal to AC

BD to AC = 1 to root 3

The bottom of p-abcd is square, the bottom of PD ⊥ is ABCD, and the point E is on the edge Pb,
(1) Verification: plane AEC ⊥ plane PDB
(2) If e is the midpoint of Pb, whether there is a point F on the edge PC (excluding the endpoint) so that the plane AEC of DF ‖ exists, find out the position of point F, if not, explain the reason
The specific process, mainly the second question and urgency! The additional wealth value of the accurate answer given to me tonight is 30

(1) AC ⊥ PD. (≁ PD ⊥ bottom ABCD) AC ⊥ BD, ≁ AC ⊥ plane PDB, AC ∈ plane ace. ≁ plane ace ⊥ plane PDB, and (2) if the center of the bottom is O., then OE ∥ DP (median line), DP ∥ plane AEC (∥ o ∈ AC) if there is a point F on the edge PC (excluding the end point), making DF ∥ plane AEC, then the plane PDF ∥ is flat

As shown in the figure, in diamond ABCD, ab = 5, ∠ BCD = 120 °, then the length of diagonal AC is______ ．

∵ AB = BC, ∵ B + ∵ BCD = 180 °, ∵ BCD = 120 °, ∵ B = 60 ° ∵ ABC is equilateral triangle ∵ AC = AB = 5, so the answer is: 5

As shown in the figure, in diamond ABCD, ab = 5, ∠ BCD = 120 °, then the length of diagonal AC is______ ．

∵ AB = BC, ∵ B + ∵ BCD = 180 °, ∵ BCD = 120 °, ∵ B = 60 ° ∵ ABC is equilateral triangle ∵ AC = AB = 5, so the answer is: 5

As shown in the figure, in diamond ABCD, ab = 5, ∠ BCD = 120 °, then the length of diagonal AC is______ ．

∵ AB = BC, ∵ B + ∵ BCD = 180 °, ∵ BCD = 120 °, ∵ B = 60 ° ∵ ABC is equilateral triangle ∵ AC = AB = 5, so the answer is: 5

As shown in the figure, in diamond ABCD, ab = 5, ∠ BCD = 120 °, then the length of diagonal AC is______ ．

∵ AB = BC, ∵ B + ∵ BCD = 180 °, ∵ BCD = 120 °, ∵ B = 60 ° ∵ ABC is equilateral triangle ∵ AC = AB = 5, so the answer is: 5

It is known that the circumference of the diamond ABCD is 8cm, ∠ BCD = 120 ° and the diagonals AC and BD intersect at point O. the lengths of AC and BD are calculated

The perimeter of ∵ rhombic ABCD is 8cm, ∵ BCD = 120 °, ∵ AB = BC = 2cm, ∵ ABC = 60 °, ∵ ABC is equilateral triangle, ∵ AC = 2cm. ∵ AC and BD are bisected vertically, ∵ OA = 1cm. ∵ ob = 22 − 12 = 3cm. ∵ BD = 23cm

As shown in the figure, in diamond ABCD, e and F are the points on BC and CD respectively, ∠ B = ∠ EAF = 60 ° and ∠ BAE = 20 ° to find the degree of ∠ CEF

As shown in the figure, connect AC. in diamond ABCD, ab = BC, ∵ ∠ B = 60 ° and ∵ △ ABC is equilateral triangle, ∵ AB = AC, ∵ BAE + ∠ CAE = ∠ BAC = 60 ° and ∵ CAF + ∠ EAC = ∠ EAF = 60 ° and ∵ BAE = ∠ caf, ∵ B = ∠ ACF = 60 ° in △ Abe and △ ACF, ∵ B = ∠ acfab = AC ∠ BAE