Simple math and geometry problems in the second year of junior high school 15

1) Proof of: CEF=90°

aef+∠afe=∠aef+∠ced=90°∴∠afe=∠ced

a=∠d△aef∽△dce

ef/ceaf/de

ae=deef/ce

af/ae∠a=∠fec

AEF ECF (both sides are proportional, the angles are equal) 2) Let ab bc=k, whether there is such a k value, obtained from (1).

Angular EFC = Angular EFA

Because the angular EFC is not a right angle.

So the angle EFA cannot be equal to the angle FCB

If AEF is similar to BFC.

Then the angle CFB = angle EFC = angle EFC = angle EFA = 60 degrees.

Let af=abc=2ae=2 3a

fb=ab=3ak=ab/bc=√3/2

Because ab=ac

So b= c

Because b=2 a

So 5 a = 180°

a=36°

Angle 2 plus angle 3 is 90 degrees, and angle 1 is 45 degrees, so the sum is 135 degrees.

Let the square side length = a

ac= T-shirt 2a

ac/ce=cd/ac=√2/2

Pyrocavity ACD ECA (SAS).

2=∠cae

1=∠cae+∠3

1 + 2 + 3 = 90°

Then the area of ABC is AC*BC 2=24 then CE=24*2 AB= In trapezoidal ABCD, AB Cd, D=90° DA=CE= (the distance between parallel lines is equal everywhere) In RT ACD, CD = AC -AD =6 Pythagorean theorem) CD=

Find the angle of a according to the cosine theorem. Then according to s=.

Proof: Extend the de intersection bc to f because the quadrilateral abcd is a parallelogram.

So ad bc ab=dc

Because de ad

So de bc

So DFC=90°

And because ecb=45°

So EFC is an isosceles right triangle. So ef=fc is in bef and dfc.

ebc=∠cdf

efb=∠dfc

ef=fc, so bef dfc

So be=dc=ab

The intersection point from de to bc is k, and the intersection from be to dc is h. According to the angle relation, it can be obtained that the angle Kdc is equal to the angle EBC, according to the inverse theorem, CE is the bisector of the angle C, and the triangle EDC is all equal to EBC, and the answer is DE.

Extending De to F, ECB=45, De AD, EFC is an isosceles right triangle, EF=CF, BFE and DFC congruence, BE=DC=AB

The conditions are not right, the vertical relationship is not suitable for the original.

Unless ABCD is a parallelogram, then be ec.

Because AD is parallel to BC and DE is perpendicular to BC. The extension wire of the DE intersects F with BC, and the result is easy to obtain.

Pictures will not be sent. You can draw while looking.

Let the right side of the small square be ab

The lower side is BC

The intersection below the small square and the big square is d

The intersection point on the right is e

Do o to ab the perpendicular line intersects at point f

Then do the perpendicular line from O to BC to intersect at the point G

in god and foe.

This liter god= foe

go=eoogd=∠oef

So god is all equal to foe

Rotate the foe to god

This four-year-old square is the 1 4 of the Little Four

1 When O is the midpoint, OAB OCD

2 The shaded part is

Answering process. Since DF parallels BC, it is obtained, BCD= CDE, and, CD bisects ACB, gets Annihilation or Out, BCD= ACD, and CVE= ACD, DE=CE

In the same way, suppose that BC extends the line CG, since talking about this DF parallel to BC, it is obtained, CFE= FCG, and CF is bisected with ACG, and CFE= FCG, and CFE= SFE= SFE, EF=CE

de=ef .