E is a point on side AD of a rectangle ABCD so that DE = 6. Also DA = 8, DC = 6. If CE is extended meets circumcircle of rectangle at F, find length of DF & FB.
Answers
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.
Answer:
5√2
Step-by-step explanation:
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 .