Math and Geometry Problems... It s harder!

It's not difficult, the first question you have already solved, I will not solve it here.

The second question, connecting oa, ob, passing the point o as ab perpendicular line intersection ab to c, then you can get oc=1 immediately, in the direct triangle oac, oa=2, oc=1, so you can find the angle aoc = 60 degrees, so there is an angle aob = 120 degrees, the rest is the same as the first question.

The third question is simpler, and it can be proved that the triangle OAB is an equilateral triangle, so the angle AOB = 60 degrees is obtained, and the connection oo is connected', then oo' bisects ab perpendicularly, and the right triangle can be solved to find it. The rest is begging for yourself.

1) 1:3 [Visually you have made it].

Analysis] Let the midpoint of the original inferior arc ab be m

Connect OA, OB, OM, MA

then om=oa=ma

OMA is an equilateral triangle.

aom=60°

aob=120°

Arc ab = 120° 360° 2 2

= 90 degrees: 270 degrees = 1:3.

Angle AOB = 120 degrees, AB arc length = 4 degrees 360 degrees = 4.

oo'=2 root number 3.

2) 4 3 over the O point for oh ab, h for the vertical foot. Because OH = r = 1 and OA = r = 2 oh, the angle AOH = 60° and the angle AOB = 120°, so the inferior arc ab = 4 3

3)2 3 ab=r, so aob, ao'b are all equilateral triangles, so oo'=2√3

As shown in the figure, in RT ABC, ACB=90°, AC=4 under the root number 3, the circle O with AC as the diameter intersects AB at the point D, the point E is the midpoint of BC, and the ob and de intersect at the point F.

Find the value of ef:fd.

t=1297772130421

The trapezoidal is isosceles trapezoidal b c eaf 60°, dc ab ad 6, ad bc

bad adc 120°, bae fad 120° 60°,df 2cf,dc 6, de 4,cf 2,af =ad +df -2ad*df*cos adc,af =36+16-2*6*4*(-1 2)=76,af= 76,abe is rotated counterclockwise 120° with a as the center, ab coincides with ad, e point is m, ab ad, b adm 60°, adf adm 180°, fdm in a straight line, dam bae, maf dam fad 60°, maf eaf 60°, am ae, af af, aef amf, ef mf md fd, let be md x, ae am a, maf adm 60°, m is the common angle, afm dam,a (x+4)=x a,,x a=6 76, the solution is x= ,ef=4+

I have a way to use analytic geometry.

For convenience, fold the graph so that A is in the lower left corner, A is the origin, and AD is in the +x direction.

d(6, 0)

The abscissa of c = AD + BCCOS60 = 6 + 6COS60 = 9

The ordinate of c = bcsin60 = 6sin60 = 3 3

c(9, 3√3)

Abscissa of f = AD + BFCOS60 = 6 + 4cos60 = 8

The ordinate of f = bfsin60 = 4sin60 =2 3

The slope of af p = tanbaf = 2 3 8 = 3 4

Slope of ae k = tanbae = tan(baf + eaf) = tan(baf + 60) = (tanbaf + tan60) (1 - tanbaf*tan60).

Equation for ae: y = 5 3x

Take y = 3 3 (ordinate of c), x = 3 5

e(3/5, 3√3)

ef² = (3/5 - 8)² 3√3 - 2√3)² = 1444/25

ef = 38/5

Make a perpendicular line at point F and cross EC at point H, you can find that be=ab*cos60=6*. So CE = 3 + 6 = 9 cm, CF = 1 3CD = 2 cm, CH = 1 cm, EH = 8 cm, FH = 3 cm, according to the Pythagorean theorem of the right triangle: the square of EF = 64 + 3 = 67, EF = root number 67

Child.. We won't hit you ...

Passing the point D as the parallel line of CE intersects BG at H, and then the corner edge is used to prove DHG ECG

Dg=eg can then be demonstrated

If you pass the point E to make a parallel line of BC, extend AB and have an intersection point with the parallel line, which proves that B is the midpoint, so G is the midpoint.

It's not the same as the one above!

Passing the point D as the parallel line of CE intersects BG at H, and then the corner edge is used to prove DHG ECG

Dg=eg can then be demonstrated

Is the O point the midpoint of AB?

If so, the perpendicular line of the BC and AC sides is made at the O point, and the perpendicular foot is E and F.

Triangle ODM is similar to OEN.

om/on=od/oe=ce/oe

If o is the midpoint.

om：on=k

If not, and bo:oa=a, then om:on=ak

Hello, would love to help you.

What is the point in the question?

Let's add it.

Feel free to ask.

o It should be the midpoint, right?

om=k×on

It's simple. You can guess it, you move the M point, so that OM and AC are parallel, you can guess this relationship, and then generalize, let M be any point on BC, and prove it.

It is very easy to make perpendicular lines from point o to the two right-angled sides, and two triangles come out. Finally, it is calculated according to the trigonometric function relationship and the line segment relationship. It's more cumbersome, and you have to find a relationship slowly.

In the right triangle ABC, the angle ACB=90°, AC=5, BC=20 3, rotate the triangle ABC counterclockwise around the point C (0 180°) to obtain the triangle A'b'c, when the triangle is aa'When the area of c is equal to 25 3 4, the rotation is stopped, and the line AB intersects the line A'b'For p, find the length of AP.

(1) Pass the point C to do the de perpendicular line ch to hand over the de to H

It is known by ce=xab=cd=2

de=root number(4+x 2).

CH=2x DE from the area of the triangle cde

Obtained by ch bf = ce be.

2x root number (4+x 2) bf=x (x+2) gives bf=y=2(x+2) root number (4+x 2) and bf de, intersection de at point f, bf and edge cd intersection at point g, then there is x maximum of 2 (f coincides with point d) 0

Fold along AC, point D coincides with point E, so AD=AE, CE=CD, angle ACE=angular ACD (axisymmetric, corresponding to the edge, corresponding to the equal angle).And because the quadrilateral ABCD is trapezoidal, AD BC, so the angle ace = angle CAD (the two parallel lines are equal in the wrong angle), so the angle ACD = angle CAD, then AD = CD (the same triangle is equal to the angle), so AD = CE = CD = AD, so the quadrilateral AECD is a diamond.

Find the area: You can't find the height, no way.

It can't be isosceles because ab=ac

d is the midpoint of the edge bc.

If the connection EFEF is a midpoint, it will be parallel.

Triangle median line) is fine if its side EF is not parallel to BC, but it is definitely not an equilateral triangle.

So EF and BC will not be parallel.

Thanks for adopting!

Proof: Extend Fe to pass Ba and Cd to P and Q respectively, take the AC midpoint M, and connect EM and FM because E is the midpoint of AD and M is the AC midpoint.

So EM is the median of ABC.

So em ab 2 and me ab

The same is true for FM CD 2 and MF CD

Due to ab cd

So me mf

So MEF MFE

Because of me ab

So ape mef

Because of MF CD

So CQE MFE

So ape cqe

Because of EF gh

So ape pgo 90°, cqe qho 90° so pgo qho

i.e. AGH DHG

The answer is correct, but if you turn the "e" and "f" upside down, you swap the "e" and "f" in the answer, and everything is clear.