In the hexagon ABCDEF, ∠ a = ∠ D, ∠ B = ∠ e, ∠ C = ∠ F, try to guess the position relationship between AF and CD, and explain the relationship

The sum of extended CD, which intersects with the extension line of AB and Fe at the inner angles of G and H hexagons is 180 ° * 4 = 720 °∠ a + ∠ B + ∠ C + ∠ D + ∠ e + ∠ f = 720 ° because ∠ a = ∠ D, ∠ B = ∠ e, ∠ C = ∠ F, so ∠ B + ∠ C + ∠ d = 720 ° / 2 = 360 °∠ g = 180 ° - ∠ GBC - ∠ GCB = 180 ° - - ∠ C) = ∠ B

If the base of p-abcd is an entire hexagon, and the vertical plane of PA is ABC, PA = 2Ab, then the relationship between Pb and BD is as follows:

Because PA is perpendicular to the bottom, AB is the projection of PA on the bottom. Because AB is perpendicular to BD, Pb is perpendicular to BD. the angle between Pb and BD is 90 degrees

As shown in the figure, PA ⊥ plane ABC, plane PAB ⊥ plane PBC, verification: ab ⊥ BC

It is proved that as shown in the figure, a is used to make ad ⊥ Pb in D,
∵ plane PAB ⊥ plane PBC, plane PAB ∩ plane PBC = Pb, ad ⊂ plane PAB,
⊥ plane PBC,
And ∵ BC ⊂ plane PBC,
∴AD⊥BC，
And ∵ PA ⊥ plane ABC, BC ⊂ plane ABC,
∴BC⊥PA，
And ∵ ad ∩ PA = a,
ν BC ⊥ plane PAB,
And ∵ ab ⊂ plane PAB,
∴BC⊥AB

As shown in the figure, P is a point outside the plane of △ ABC, and PA ⊥ plane ABC. If O and Q are the perpendicular centers of ⊥ ABC and ⊥ PBC respectively, it is proved that OQ ⊥ plane PBC

It is proved that ∵ o is the vertical center of △ ABC, ᙽ BC ⊥ AE. ∵ PA ⊥ plane ABC, and BC ⊥ PE is obtained according to the three perpendicular line theorem
⊥ plane PAE. ∵ q is the vertical center of ⊥ PBC, so if q is on PE, then OQ ⊂ plane PAE, ∵ OQ

If point P is a point outside the plane of △ ABC, PA, Pb and PC are perpendicular, and Po ⊥ plane ABC is at point O, then o is () A. Exocentrism B. Heart C. Be attached to one's heart D. Center of gravity

The results showed that: Ao was connected and prolonged, BC and D were connected with Bo and AC and E were crossed;
Because PA ⊥ Pb, PA ⊥ PC, PA ⊥ faces PBC, so PA ⊥ BC;
Because Po ⊥ is ABC, so Po ⊥ BC, so BC ⊥ is Pao,
So Ao ⊥ BC is ad ⊥ BC;
Similarly, be ⊥ AC;
So o is the perpendicular center of △ ABC
Therefore, C

Take two points a and B on the straight line m so that ab = 10cm, and then take a point P on m so that PA = 2cm, m and N are the midpoint of PA and Pb respectively. Find the length of line Mn

As shown in the figure, (1) when point P is on the line segment AB, Pb = ab-pa = 8cm, m and N are the midpoint of PA and Pb respectively, ﹥ Mn = PM + PN = 12ap + 12bp = 1 + 4 = 5 (CM); (2) when point P is on the extension line of line Ba, Pb = AB + PA = 12cm, m and N are the midpoint of PA and Pb respectively,  Mn = pn-pm = 12bp-12ap = 6-1 = 5

Let p be a point on the line AB, Mn is not the midpoint of PA and Pb, ab = 20cm, find Mn

MN=MP+PN=0.5AP+0.5PB=0.5(AP+PB)=0.5AB=0.5*20=10(cm)

Take two points a and B on the straight line m so that ab = 10cm, and take any point P on m so that Pb = 4cm, m, n are the midpoint of PA and Pb respectively, and find the length of line Mn D

5cm

If the point P is outside the line AB and PA = Pb, then Mn is the vertical bisector of line ab If we make a straight line through P, Mn is just parallel to ab

Of course, that is not true
PA = Pb, indicating that point P is on the vertical bisector of line AB, but there are countless lines passing through point P, that is to say, the line Mn passing through point P is not necessarily the one perpendicular to AB, so this statement is not correct

Given that the line AB = 100cm, M is the midpoint of AB, P is on the line MB, n is the midpoint of Pb, and Nb = 12cm, find the length of PA

N is the midpoint of Pb, and Nb = 12cm
Then Pb = 2Nb = 24cm
Then PA = ab-pb = 76cm

In the hexagon ABCDEF, ∠ a = ∠ D, ∠ B = ∠ e, ∠ C = ∠ F, try to guess the position relationship between AF and CD, and explain the relationship