Circle Question

s3a · #1 $Old$ December 3rd, 2008, 08:29 PM

I don't know what to do on this question and would appreciate detailed help.
Thanks in advance!

Shyam · #2 $Old$ December 3rd, 2008, 10:00 PM

Quote:

Originally Posted by s3a $View Post$

I don't know what to do on this question and would appreciate detailed help.
Thanks in advance!

triangles ADB and BDC are similar.

so,

$\frac{AD}{BD}=\frac{DB}{DC}$

$\frac{AD}{12}=\frac{12}{5}$

$AD=\frac{144}{5}=28.8\;\;cm$

$AC=AD+DC=28.8+5=33.8 \;\;cm$

$ED = AC-(14+5)=33.8-19=14.8\;\;cm$

$EC=AC-AE=33.8=14=19.8\;\;cm$

In triangle EBD,

$EB^2=BD^2+ED^2$

$EB=\sqrt{(12)^2+(14.8)^2}$

EB = 19.05 cm

NOW,

$FE \times EB=AE \times EC$

$FE \times 19.05=14 \times 19.8$

FE = 14.6 cm

got it ???

s3a · #3 $Old$ December 4th, 2008, 01:44 PM

Thank you very much! I copied and analyzed your work for my assignment but even though the assignment is over with, I'd just like to ask one little question if you can still answer:

How do you prove that triangles ADB and BDC are similar?