Reward score: 20

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Question time 20 10-5- 15

20: 1 1

In the triangle ABC, the angle BAC=90 degrees, AD is the height on the side of BC, E is the moving point on the side of BC (not coincident with points B and C), EF is perpendicular to BC, EG is perpendicular to AC, and the vertical feet are F and G respectively.

(1) verify eg/ad = CG/CD;

(2) Connect FD and DG. Please judge whether FD and DG are vertical. If vertical, please give proof;

(3) When AB=AC, is the triangle FDG an isosceles right triangle? Why?

Respondents:

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20 10-5- 16

08: 13

Is there a problem with the topic? EF is perpendicular to BC, EG is perpendicular to AC, and the vertical feet are f and g respectively. Since EF is perpendicular to BC, how can the vertical foot be F?

The answer to the first question is:

Proof: proving EG/CG=AD/CD is equivalent to proving EG/AD=CG/CD.

According to the known conditions, △EGC∽△BAC's: CG/CA=EG/BA is being converted into CG/EG = AC/BA.

Then the AC/DC=BA/AD of △BAC∽△ADC is converted into AC/BA = DC/AD. ②.

CG/EG=DC/AD can be obtained from ① and ②, that is, EG/AD=CG/CD.

The second question: ellipsis.

Respondents:

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20 10-5- 16

08:42

Respondents:

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20 10-5-25

18:08

Respondents:

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20 10-5-3 1

Eight twenty _ six p _ m

Hmm. How interesting

That's what I did.

(1) Angle ADC= Angle EGC= 90°, Angle C = Angle C, so the triangular ADC is similar to the triangular EGC, so EG/AD=CG/CD.

(2) Because Angle B= Angle B, Angle A= Angle BDA = 90°, Triangle BAD= Triangle BCA, so BA/BC=AD/CA, so BA/AD=BC/AC, because Angle B+ Angle BAD = 90, Angle BAD+ Angle DAC = 90, so Angle B= Angle DAC.

(3) Because AB=AC, angle BEF= Angle A, and angle B= Angle B, triangle BEF is similar to triangle BCA, so BD=AD, so triangle BFD congruent triangles AGD, so FD=DG, because FD is perpendicular to DG, so it is an isosceles right triangle.

Respondents:

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20 10-6-2

10: 04 p.m.

four

20 10-5-24

three

20 10-5- 15

222

2009-2-4

47

2008-4-30

10

20 10-6-2 1

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