E is a point on side AD of rectangle ABCD so that DE = 6, while DA = 8 and DC = 6. If CE is extended to meet the circumcircle of the rectangle at point F, find the length of DF. Also find the length of FB.
Answers
Answered by
1
Step-by-step explanation:
Answer:
Note that AE = AD - ED=8-6=2
Since triangle CDE is a right triangle with DC = ED=6, the pythagorean theorem yields
EC = 6 sqrt (2)
When 2 chords intersect dividing the chords into lengths of w and x, and y and z respectibely we know that wx=yz. Since AE-2, ED =6, and CE = 6 sqrt 2 we have that EF ( 6 sqrt 2) = 2(6) or rather EF = sqrt 2. Note that AC is the hypotenuse of right triangle ADC so that AC = sqrt (6^2 + 8^2) = 10. Now triangle FED is similar to triangle AEC so that sqrt 2/2 = FD / 10. Then FD = 5 sqrt 2.
Similar questions