Senior 1 Geometry Problems, Senior 1 Geometry Math Problems

Find the midpoint of PD, denoted as F, and connect AF and EF

E is the midpoint of the PC.

EF cd and CD=2EF

CD AB and CD = 2AB

EF AB and EF = AB

ABEF is a parallelogram.

be//af

and be inside a flat pad.

be planar pad

Proof is made as the midpoint F of PD, connecting EF, AF

E, F, are the midpoints of PC and PD, respectively.

ef//=1/2cd

cd//ab ,cd=2ab

ab//=ef

The quadrilateral ABEF is a parallelogram.

be//af

AF belongs to planar PAD

be planar pad

Proof: Take the midpoint F of the PD and connect EF and AF

E is the PC midpoint and F is the PD midpoint.

In PCD, EF is the median line.

EF cd and EF = 1 2CD

In trapezoidal ABCD, AB CD and AB=1 2CDAB EF and AB=EF

The quadrilateral ABEF is a parallelogram.

af be is inside the planar pad and be is outside the planar pad.

be planar pad

Make the midpoint F of the CD and connect EF and BF

Because E is the PC midpoint and EF is the triangular PCD median.

So EF PD

Because cd=2ab

So df=ab

And because of ab cd, that is, ab df

So bf ad

EF PD, BF AD are obtained

So planar bef planar apd

be on planar bef.

So be planar apd

Take the midpoint F of PD and connect EF and AF

E and F are the midpoints of PC and PD, respectively.

fe cd and ef = 1 2cd

and ab = 1 2cd and cd ab

EF AB and EF = AB

The quadrilateral afeb is a parallelogram.

be//af

AF is in the face pad, and be is not in the face pad.

Be face pad

Take the midpoint F of PD and connect AF and EF

Points E and F are the midpoints of PC and PD, respectively.

EF cd and EF = 1 2CD

ab cd and ab = 1 2cd

ef‖ab，ef=ab

The quadrilateral ABEF is a parallelogram.

Be AFBE does not belong to a planar pad, AF belongs to a planar PAD

be planar pad

Find the midpoint F on the straight line CD and connect EF and BF. Easy to know EF PD.

dc ab, df=ab, then abfd is a parallelogram, then bf ad.

So, planar bef adp.

Then be a planar pad. The original question is proven!

Make a little f on the pd, so that pf=fd, connect af and ef

EF CD and CD AB can be obtained, so EF ABE is the PC midpoint and F is the PC midpoint. So cd=2ef=2abab is parallel and equal to ef, so abef is a parallelogram, so af beaf is on a planar pad.

So be flat pad

Take the midpoint M of the PD and connect it to EM

So EM is the median line of the triangular PCD, EM parallel = 1 2cd because CD is parallel to AB CD = 2AB, EM parallel = 1 2CD = ab so EM parallel = ab

The parallelogram abem, get be parallel to am

Because AM belongs to the face pad

So be parallel to the face pad

Certificate: Do EF||PD, intersect CD to F, connect BF

e is the midpoint of the PC, ef||pd

f is the midpoint of cd, and cd=2ab

de=ab, and de||ab

The quadrilateral abfd is a parallelogram.

bf||ad

EF,BF intersect F,PD,AD versus D

and de||ab ，ef||pd

Planar efb||Flat pad

be||Flat pad

Whose proof will work? Speechless aq me, mn db, also cannot be proven.

There is a problem with this topic, ABCD is a quadrilateral.

Not a triangle.

20 2-12 2) is half the length of one of the sides of the cross-section, so the cross-sectional area s=15*2*16=480

The straight line l:mx-y+1-m=0 is obtained over the fixed point (1,1), and this point is within the circular equation, so it is verified.

2) When m=0, the equation of the straight line l is: y=1, the coordinates of m are (0,1), when m is not 0, it is easy to obtain the linear equation of the center of the circle (0,1) and m(x,y) is: y=-(1 m)*x+1, and the equation of m solved by simultaneous mx-y+1-m=0 is (y-1) 2=-x*(x-1), which is simplified to obtain (y-1) 2+(

The trajectory of the total m is a circle.

3) You can do it on the basis of the second question, there is pb=pm+mb, pb=2*pa, you can get pm=ma 3, and in the triangle cma, cm=,ca=5, you can find the length of ma, and then find pm, let m(x,mx+1-m), and p(1,1), you can find the value of m.

To draw a picture ... For this, find a point on the bottom surface at random, and then pass p to make pe a1d1 to a1b1 to e, pass p to do pf a1b1 to a1d1 to f, connect af and ae

Then let aa1=a, pf=b, pe=c

Then there are a lot of right triangles, find the angles we need a= paa1, b= apf, c= ape and then calculate the cos value in the right triangles.

For example, in APA1, cos paa1=a root number (a 2 + b 2 + c 2).

Then by analogy, cos apf=b root number (a 2 + b 2 + c 2) cos ape = c root number (a 2 + b 2 + c 2), so cos 2a + cos 2b + cos 2c = 1

(1) Seek the height of the side first. Let the side height be h, then the question is set: 4 (3+6)h 2=9+36

=>h=5/2.(2) Then the height of the platform is obtained by the Pythagorean theorem, and the height of the platform is set to hThen h + (3 2) = h

=>h²=(25/4)-(9/4)=4.∴h=2.That is, the height of the platform is 2.

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