Geometry and Mathematics Junior High School Urgency You don t have to answer after 1 o clock today

Because EF SD, in the triangle ABC, the S triangle AEG = 1 3Si quadrilateral EBCG, so E, G, F are the midpoints of AB, AC, AD, respectively. And because of cf ab, cf ab = 1 2

What can be known is: s aeg=(1 4)s abc, aeg abc, ae ab=1 2, ag ac=1 2, af ad=1 2, cf is the midline on the hypotenuse of rt acd, cf ad=1 2,; As for CF AB ......I don't know.

Only the midpoints E and F can not be determined, so the length of C and F cannot be determined.

Skillful handwriting and poor math are the foundations of the basics.

1. Solution: ABC is an isosceles right triangle, BAC 90°, AB Accd AE, AF BF

bfa＝∠cda

baf＋∠fac＝90°，∠baf＋∠abf＝90°∠dac＝∠abf

In BFA and ADC.

abf＝∠dac

bfa＝∠cda

ab＝ac△bfa≡△adc

bf＝ad，af＝cd

ad＋df＝af

bf＋df＝cd

<>1, bf, cd, df The relationship between the three is: cd=bf+df.

Proof that abc is an isosceles right triangle, ab=ac, 1+ 3=90°.

cd ae, 2+ 3=90°, while 1+ 3=90°, 1= 2In the right triangle abf and the right triangle cad, 1 = 2, ab=ac, cad abf (right angled edge, hypotenuse) cd=af, ad=bf

and af=ad+df, cd=bf+df (equivalent substitution).

ac=ae，∠cad=40°， ace=∠aec=(180°—40°)÷2=70°，∠dce=∠acd—∠ace=90°—70°=20°。Finished!

Proof: Take the n point on BC so that cn=AM, connect an in ACN and EAM.

AC=EA, CN=AM is known

Because AHC=90 degrees.

So NCA+ CAH=90 degrees.

again mae+ cah=180-90=90 degrees, so mae= nca

So the ACN congruent eam (corner edge).

So em=an (*.)

The same can be said for abn= dam

Hence bac= adm+ aem

and can= aem

Therefore ban= adm

Known ad=ab

So abn congruent dam (corner corner).

So an=dm

So we get em=dm from the equation (*).

i.e. m is the midpoint of de.

Analysis: Graphically, the relationship is clear if this is the case with DAE CDE. DE square = ae ce can not break through left and right, it seems that the auxiliary line has to be added.

There is a way to add auxiliary lines, and combined with this problem and that 2x relationship, we should easily think of the median line of the triangle.

Therefore the extension of CD and BA to F

Easy to prove cd=df

It turned out that DE turned out to be the center vertical line of CF.

Fe=CE is high in RT FDE, and now the conjectures are verified.

Proof omitted.

DF perpendicular to CB, perpendicular foot to F, CE to G, Dh perpendicular to CE, perpendicular foot to H.

It is easy to know that the quadrilateral abfd is a rectangle, bf=ad, by bc=2ad, so cf=ad=bf, and df ab, so cg=eg, rt triangle cde, dg=ce 2=cg, thus the angle cdg = angle dcg, easy to know the angle edh = angle dcg, so the angle edh = angle cdg.

and the angle cde = angle adf = 90 degrees, the angle ade = angle cdg = angle edh, so that the triangle ade is completely equal to the triangle hde, eh = ae, the triangle cde is similar to the triangle dhe, so that the square of de = eh ce, that is, the square of de = ae ce

Because the angle cde + angle cbe = 90 degrees + 90 degrees = 180 degrees, b, c, d, e are on a circle with ce as the diameter, as df perpendicular to cb, perpendicular to f, cross ce to g, it is easy to know that the quadrilateral abfd is a rectangle, bf = ad, by bc = 2ad, so cf = ad = bf, df is the perpendicular line of cb, g is the center of the above circle, gc = gd = ge, angle cde = angle fde = angle aed, dae cde, de square = ae ce

The square of de is equal to ae times ce

1.If the auxiliary line is extended by CD and BA and intersected with one point, it is denoted as point F, then DA is the median line of the triangle FCB, DA is parallel and equal to one-half BC, and the triangle FDA is similar to the triangle FCB, and the ratio of each line segment is one-half.

2 easy triangle fde is all equal to triangle cde, angle fed is equal to angle ced, because angle dae is equal to angle cde, so triangle dae is similar to triangle cde, so de ce is equal to ae de, so ae, de, ce are the square of de is equal to ae times ce