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27 problems in junior high school mathematics
27. Solution: Because OA OA, OB is the two roots of the equation x 2-8x+ 16 = 0.

So OA=4

OB=4

Because the straight line y=kx+b intersects the y axis, and the x axis intersects a and b.

Therefore, let a (0 0,4) b (-4,0) generation y=kx+b obtain:

k= 1

b=4

Replace k= 1 b=4 with y=kx+b: y=x+4.

So the analytical formula of the straight line y=kx+b is y=x+4.

28. (1) It is proved that the intersection D makes DH perpendicular to AB and connects GH.

So AHD angle = BHD angle =90 degrees.

Because DE is perpendicular to BC and e.

So the angle bed =90 degrees.

So angle BHD= angle bed =90 degrees.

Because BD bisects the angle ABC

So angle ABD= angle CBD.

Because BD=BD

So triangular BHD and triangular bed congruence (AAS)

So BH=BE

Because angle ABD= angle CBD

Blood sugar = blood sugar

So triangle BHG and triangle Berg are congruent (SAS).

So angle BGH= angle BGE.

Because AG is perpendicular to BD and g.

So AGD angle =90 degrees

So AHD angle +AGD angle = 180 degrees.

So a, h, d and g are four-point circles.

So angle BAD= angle BGH

So Angel Budd = Angel ·BGE

So a, b, f and g are four-point circles.

So angle ABD= angle AFG

So angle CBD= angle AFG

Because angle AFG+ angle DFE= 180 degrees.

So angle DFE+ angle CBD= 180 degrees.

(2) Solution: Because BD bisects the angle ABC.

So angle ABD= angle CBD= 1/2 angle ABC.

Because the angle ABC=90 degrees

So the angle ABD= the angle CBD=45 degrees.

Because AG is perpendicular to BD and g.

So AGB angle =90 degrees

Because the angle ABD+ the angle AGB+ the angle package = 180 degrees.

So angle ABG= angle ABD=45 degrees.

So AG=BG

From Pythagorean Theorem:

AB^2=AG^2+BG^2

AB= root number 2*BG

Because CG is perpendicular to BC

So the angle BCG=90 degrees

Because angle BCG+ angle CBD+ angle BGC= 180 degrees.

So the angle CBD= the angle BGC=45 degrees.

So BC=GC

From Pythagorean Theorem:

BG^2=BC^2+GC^2

BG= root number 2*BC

So AB/BC=2

Because virtue is vertical BC.

So angle bed = angle DCE=90 degrees.

So angle ABC+ angle bed = 180 degrees.

So parallel AB

So DE/CE=AB/BC

So DE/CE=2

CE= 1/2DE

In the right triangle DEC, it is obtained by Pythagorean theorem:

CD^2=DE^2+CE^2

So CD= root number 5*CE

Because angle DEB= angle BCG=90 degrees

So DE parallel CG

So DE/CG=BE/BC=ME/HC=BE/CE.

Angle FDM= Angle FCH

Angle FMD= Angle FHC

So the triangle DFM is similar to the triangle CFH (AA)

So DM/CH=DF/CF

Because angle DFE= angle CFG

Angle FDM= Angle FCH

So triangle DEF is similar to triangle CGF (AA)

So DE/CG=DF/CF

So DM/CH=EM/CH.

So DM=ME= 1/2DE.

So DM=CE

So CD= root number 5*DM