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Mathematics 21 Online
OpenStudy (anonymous):

Find AF:FC in the attached geometrical figure.

OpenStudy (anonymous):

OpenStudy (anonymous):

Can u do doc instead of docx or pdf...

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

tks.

OpenStudy (anonymous):

OpenStudy (akshay_budhkar):

i believe the answer is 1:1 because ad:db=af:fc and ad:db=1:1.. n i can prove that

OpenStudy (akshay_budhkar):

its a property i believe

OpenStudy (akshay_budhkar):

estudier?/???

OpenStudy (anonymous):

I just downloaded it, looking now....

OpenStudy (akshay_budhkar):

i gotta go mkuma.. but i still believe it is right.. estudier will help you thiugh.. sorry..

OpenStudy (anonymous):

thanks ...!

OpenStudy (anonymous):

Do I have to show ABC is 90 first?

OpenStudy (anonymous):

Or I can just assume?

OpenStudy (anonymous):

Blasted censor!

OpenStudy (anonymous):

ASSUME

OpenStudy (anonymous):

We can't assume it....if we can get by some property then it is

OpenStudy (anonymous):

OK, it is, I will include it..

OpenStudy (anonymous):

Let GBC = theta = BED = DEA and BDF/DBC are right. So ADE/DEB are congruent 1,1,2 type triangles and AD:DB is 1:1 and then as Ashkar says.

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