Mathematics and Geometry in Senior 1 It is best to explain in detail, to ask the first question, not

It should be ab1=b1c= 2a, right?

1)∵ab=bb1=a，ab1=√2a

According to the Pythagorean theorem, abb1 is rt, abb1=90°ab bb1

This prism is a straight prism.

bb1⊥be

And BCE= bad=60°,bc=a,ce=1 2cd=1 2a is obtained from the cosine theorem, be= 3 2a

According to the Pythagorean theorem, BCE is rt, bec=90° and ab cd

abe=90°

ab bebe face abb1

ab1⊥be

2) Connecting AC, intersecting BE to O, connecting OF, FE, and FB only needs to satisfy of AB1, and since AB1 plane BEF is in diamond-shaped ABCD, AC bisects DAB

oab=30°

and abo=90°

bo=√3/3ab=√3/3a

and be= 3 2a

bo：be=2：3

ao：ac=2：3

When b1f:b1c=2:3, there is of ab1, and ab1 is a plane bef

Therefore, when b1f:fc=2:1, ab1 plane bef

There is a problem with the question, abb1 is not a triangle at all.

Dear, you need to use mathematical formulas, mathematical figures, I will write them for you in word

(2) Set: The midpoint coordinate is m(x,y).

With the straight line pm perpendicular to the straight line op, then:

y 0) (x 5)] y 0) (x 0)] = 1 Simplification yields the trajectory equation for point m as:

x(x 5) y =0 [must be indicated: the trajectory of point m is within the circle].

Since om mp, so m is on a circle with op as the diameter, so the coordinates of m meet x +y -5x=0, and m is in the circle x +y =16, so x<16 5

Therefore, the trajectory equation for m is: x + y -5x=0 (0 x<16 5).

PM is perpendicular to OP

m is on a circle with op as the diameter.

x(x-5)+y^2=0

x^2+y^2-5x=0

Let the equation with k (slope) be solved.

Okay lz

The (1) of this question is very difficult, and there are three difficulties.

1. The face is vertically placed obliquely, and it is extremely difficult for the reclining face to directly take the plane angle of the dihedral angle 2, and it is almost impossible to directly establish the coordinate system solution (reason: PDA is unknown, and it is a more difficult task to establish the coordinate system and force the solution. )

3. The figure itself is confusing, the right-angled trapezoidal shape on the bottom surface is actually CDA, BAD is 90 degrees, fortunately, the author of this question made a low-level mistake, and for no reason there is an E is the pb midpoint, which seems to be the condition that the village is not in the store, is the key to this question!

Through E, EF AB is made in the planar PAB, and the PA is connected to F to FD

E is the PB midpoint.

EF is the median line of PAB.

That is, ef = ab 2 = cd

CD AB again

cd efcdfe is a parallelogram.

CE DFAB Planar Pad, DF C Planar Pad AB DF

That is, AB CE

pd=ad

f is the midpoint on the bottom of this isosceles triangle PA.

According to the three-in-one, DF is the high line on the bottom edge of the DAP.

That is, DF PA

Thus CE PA

Since ab pa=a; AB, PA C Plane PAB CE Plane PAB

CE C planar PBC today

Other words. Planar PBC Planar PAB

It's very simple. The Pythagorean theorem yields ec1=eb, takes the midpoint m of BC1, and connects EM, then EM BC1

It is also easy to prove em af, and af bc, em bc em face bb1c1c

EM is contained in the surface EBC1, the surface EBC1 and the surface BB1C1C

Face a small flower drum, and draw a tiger on the drum. Mom mends it with a cloth.