Solving Geometry Problems Middle School Solving Geometry Problems for Middle School

Let the side length of the square be x, and the triangle corresponds to the column ad bp=dn pn, that is, x bp =(dm+1) 3 (1).

ad/pc=dm/mp

i.e. x (x +bp) = dm 4 (2).

The numerator (2) and the denominator are reversed and simplified.

1+bp/x =4/dm (3)

Bring (1) into (3).

1+3 (dm+1)=4 dm gives dm = 2 or -2 (rounded).

Upstairs is also, you can do that.

Hello, this question should use the square and triangle properties and area formulas, the following is the solution process:

The triangle ABE rotates 90 degrees around A and it is clear at a glance.

Invited, but disrespectful.

Take a look below and click to enlarge:

I feel that it is equal to 18, 2 isosceles right triangles, the bottom edge is 12, the triangle area is multiplied by the height of 2 is 36, and then divided by 2 is a triangle.

Extend AE to CB for point I

Then the company square becomes x

Then you can do it with similar triangles and find x

AEB is similar to AIB

The squares are bisected perpendicular to each other.

AE is half of the diagonal.

The area of the square = 2aexae = 72

Solution: Extend EC to G so that CG=AM, and connect BG

Because ABCD is a square.

So ab=bc, angle a=angle ecb== angle gcb=90 degrees.

So the triangle BCG is all equal to the triangle BAM

So angular cbg=angular abm, bg=bm

Because the angle mbe = 45 degrees.

So the angle EBG = 45 degrees.

In the triangle BME and the triangle BGE BG=BM=bm, the angle EBG=angular mbe, and BE=BE

So the triangle BME is all equal to the triangle BGE

So the area of the pentagon ABCCEM is 2 times the area of the triangle BME.

Make BH perpendicular to EM, and perpendicular to H

Then because the triangle bme is all equal to the triangle bge

So the high BH=BC on the corresponding edge

Because me=10, bc=12

Therefore, the area of the triangular BME is 1 2*10*12 60

The area of the pentagonal ABCCEM is 120

Area of the triangle MDE Area of the square ABCD – The area of the pentagonal ABCCEM

Because bh=bc

Therefore, it is not difficult to prove that the triangle BMA is congruent with the triangle BMH, and the triangle BEC is congruent with the triangle BGC.

So mh=ma, eh=ec

So ma+ec=me=10

And because am+md=ab=12, ec+ed=dc=12

So md+ed=12+12--10=14

Because the area of the triangle mde has been found 24

So md*ed=48

From md+ed=14 and md*ed=48, ed=6 or ed=8 can be obtained

When ed=6, ec=4, ed:ec=3:2 the area of the triangle EFC is 24(2 3) squared 32 3

In this case, the area of the triangle MDE + the area of the triangle CEF is 24 + 32 3 34 and 2 3

When ed=8, ec=2, ed:ec=4:1 the area of the triangle cef is 24(1 4) squared 3 2

In this case, the area of the triangle MDE + the area of the triangle CEF is 24 + 3 2 25 and 1 2

30 and 48 ec=6 when ed=6Area of the triangle EFC Area of the triangle DEM 24

At this time, the area of the triangle MDE + the area of the triangle CEF is 24+24 48

When ed=8, ec=4, ed:ec=2:1 The area of the triangle CEF is 24*(1 2) 2=6

At this time, the area of the triangle MDE + the area of the triangle CEF is 24+6 30

Hello, the answer you want is:

Because ae = Ba2 and Ba = cd

So, ae = cd2

And because of the ae cd

So, AE is the median line of the triangular FCD.

So, E is the midpoint of DF.

Whereas, the triangle FOD is a right triangle.

So. The midline OE on the hypotenuse is equal to half of the hypotenuse DF, i.e., OE = 1 2DF

Proof: Connect BF

bo=od, bof= dof, bof is all equal to dof and ba:ae=2:1, so bfd is an equilateral triangle (the landlord can prove himself, omitted) point a is the center of the equilateral triangle bfd.

fe=ed, we can know that the point e is the midpoint on the hypotenuse FD of the right triangle FOD.

be=fe=ed=1/2fd

It's you again, this question is very interesting, I'll take a look, I drew ...... by myselfMake eh fo ea ab 1 2 eh bo 1 2 i.e. eh od 1 2

EF FD 1 2 EF ED In RT, OE=EF=ED OE=1 2DF is it clear, do you want a diagram?

ce⊥cf

Reason: Because cd=ac

So ACD is an isosceles triangle.

Because F is the midpoint of AD, CE

So CF divides ACD equally

Because CE bisects the angle ACB

So the angle ace = angle ECB, the angle ACF = angle FCD so the angle ECF = 90 degrees.

So vertically.

The triangle ACD is an isosceles triangle, and CF is the angular bisector of the angular ACD, so the angular ECF is half of the angular BCD, i.e. 90 degrees.

CE is perpendicular to CF.

Perpendicular because cd=ac f is the midpoint of ad.

So the triangle ACF is all equal to the triangle DCF

So the angle ACF is equal to the angle DCF

Because CE bisects the angle ACB

So the angle ace is equal to the angle BCE

The sum of the above four corners = 180 degrees.

So the CE CFs are perpendicular to each other.

Vertical relationships.

cd=ac, f is the midpoint of ad.

acf=∠dcf

CE Split ACB

ace=∠bce

ACE + ACF = 90 degrees.

Perpendicular because: according to the properties of isosceles triangles, the angle ACF = angle DCF = 1 2ACD, and the angle ACE = angle BCE = 1 2ACB

There is a reason: the angle BCD is a flat angle.

So. Angular ACB + angular ACD = 180°

So, 1 2 angle ACB + 1 2 angle ACD = 90° angle ACE + angle ACF = 90°

So, ce, of vertical.

Certificate: cd=ac

cad=∠adc

f is the midpoint of AD, i.e., af=df

cfd≌△cfa

acf=∠dcf

Ace= BCE

acf+∠dcf+∠ace+∠bce=180∘∵∠acf=∠dcf; ∠ace=∠bce∴∠ecf=90∘

i.e.: CE CF

eb=ec=ef.

DFH+ FDH=90°, A+ ACH=90°, get DFH= ACH, get EF=EC

Even cd then 1= a= bca, d, b, c, e four points are round, so 2= 3= dbc and ecb= bca, and 4= a= bce, so eb=ec

Adoption is a virtue

Solve the outer angles of the trigonometric row = the sum of the two non-adjacent inner angles.

There is 3= 1+ 2= 4

There are 4 + 2 = 180-96 = 84 degrees = 2 2 + 2 = 3 2 to get 2 = 28 degrees.

4 = 28 * 2 = 56 degrees.

So dac=180 degrees-56*2=68 degrees.