Mathematical geometry in the second year of junior high school, mathematical geometry in junior high

1) Point E moves along the direction of A-D, point F moves along the direction of D-A, the velocity is 1cm s, so ae=df, and in the rectangular abcd, ab=cd, so abe dcf

So: quadrilateral, bcfe = quadrilateral, abcd-triangle, abe-triangle, cdf3*2-2*, triangle, abe

6-2T because the area of a quadrilateral BCFE is three-quarters of the area of a rectangular ABCD.

So 6-2t = three-quarters * 2*3 solution t=

2) Extend the BE and CF at point G

abe dcf so be=cf

and ad bc so ge be=gf cf so ge=gf so be+gc=cf+gf i.e. gb=gc because bgc=60°, gbc is an equilateral triangle.

So EBC= BGC=60°= AEB

So ae = ab*cot60° = two-thirds of the root number three.

t = 2/3 of the number three.

3) Connect BF and CE at point M

The diagonal angle between BF and CE of the quadrilateral BCFE is 90°, and RT BEM = RT cmf can be obtained

So bec cfb so bf-fm=ce-em that bm=cm so fbc=45°, as fn bc to bc at the point n so bc = bn + cn = fn + fd = cd + df = 2 + t = 3 so t = 1 just like that, if there is anything you don't understand, you can talk.

1. The other two parts are 1 4, 2*1 2*ab*ae=1 4*ab*ad, ae=3 4cm, t=

2. The angle between BE and CF is 60, indicating that the angle BCF=60, DF=DC Tan60=2 root number 3, T = 2 root number 3

3. BF is perpendicular to CE, which means that the angle ECB = angle ECD = 45, DE = 2, so AE = 1cm, t = 1

We can see that in the triangle BCE is equal to the triangle ACD. So ad=be, ed=de, ec=dc, so actually we're comparing gc and fc. Whereas the triangle EFC is equal to the triangle DGC (because the angle ECF = angle ECD = 60 degrees; ec=cd, in the triangle bce is equal to the triangle acd we can also get the angle bec = angle adc, 2 angles sandwiched on one side).

So fc=gc, so the distances are exactly equal and arrive at the same time.

From the inscription, it can be seen that the quadrilateral ABCD is a diamond with an angle of 60°, and ABC is an equilateral triangle.

aef=∠acd=60°

ap=ce

AB=BC equilateral BPE

ape=∠ecf=120°

In APE & ECF

pae=∠cef，ap=ce，∠ape=∠ecf∴△ape≌△ecf（asa）

ae=ef

Connect AF, certificate triangle ABC, AEF congruence, it's good, the target is called all written, what that 60 degrees, and pay attention to the wrong angle or something, it's all useful, it won't hi me.

(1) (2) is correct, 3) is solved, because the straight line oc is translated to the left by 4 unit lengths to obtain the straight line EF, then the analytical formula of the straight line EF is y=x+4,

Your diagram is not right, the top two vertices are ab, the bottom two are cd, and the two heights are set to h and the two highs cross cd in e and f

Let ce be x, and you can do the equation 5 -x =3 -(11-x).

You can find the height, and if you have the height, you can find the area.

(1) Because: the BOC will be rotated around the point C by 60° clockwise to obtain a triangular ADC, so that BOC is equal to ADC

It can be concluded that co=cd

And because bco+ aco=60°

bco=∠acd

So aco+ acd=60°

The two sides are equal and the angle is 60 degrees, so the triangle cod is an equilateral triangle (2) According to the first question, the triangle cod is an equilateral triangle.

So: the angle CDO is 60 degrees.

The triangle BOC and the triangle ADC are congruent triangles, and the angle ADC = angle BOC = = 150 degrees.

Angular ado = angular ACD - angular cdo = 150-60 = 90 degrees.

So the triangle AOD is a right-angled triangle.

3) If AOD is an isosceles triangle, then the angle AOD=angle ADO can be obtained to -60=360-110--60

Solution: (1) BOC ADC, OC=DC --1 minute) OCD=60°, OCD is an equilateral triangle --1 minute)(2) AOD is RT -1 minute).

Here's why: OCD is an equilateral triangle, ODC=60°, BOC ADC, 150°, ADC= BOC= 150°, ADo= ADC- ODC=150°-60°=90°, AOD is RT -2 points).

3) OCD is an equilateral triangle, cod= odc=60° aob=110°, adc= boc= ,aod=360°- aob- boc- cod=360°-110°- 60°=190°- ado= adc- odc= -60°, oad=180°- aod- ado=180°-(190°- 60°)=50°

When aod= ado, 190°- 60°, 125° -2 points) When aod= oad, 190°- 50°, 140° -2 points) When ado= oad, -60°=50°, 110° -2 points) In summary: when =110° or 125° or 140°, AOD is an isosceles triangle --1 minute).

(1) According to the title, it is 60 degrees, so the angle 0cd=60 degrees, oc=od, so the three angles are all 60 degrees, which is an equilateral triangle.

2) Because it is 150 degrees, the angle ADC=150 degrees, because the triangle OCD is an equilateral triangle, so the angle CDO=60 degrees, the angle COD=60 degrees, the angle AOC=360-110-150=100 degrees, so the angle ADO=90 degrees, the angle AOD=40 degrees, is a right triangle.

3) According to the second question relation, the angle ado = -60 degrees, the angle aod = 360-110- -60 degrees, so when the two angles are equal, =125 degrees, because the angle oad=50 degrees, so =110 degrees is also OK.

Certificates: In RT PHC and RT PDC.

hc = dc (known).

pc=pc (common edge).

rt△phc≌rt△pdc

hpc=∠dpc ∠pch=∠pdc

PCH+ PCD=45° and DPC+ PCD=90° PCH= PDC=1 2 DPC

HPC = 60° (can be a column equation, or can be extrapolated by proportion).

Extend AD so that DE is equal to HP even EC

Since hc is equal to dc, and there are two right angles, we can prove that the triangle hpc and dec are congruent.

So pc is equal to ec hpc = e= cpd = 180- pce pce= hcd = 45 degrees.

So hpc=(180-45).