△ A B C , the circle passes through A and B and intersects △ A B C at point D and E . Extend A B and D E so that they intersect at point F , connect line C F .
Given a circle andIf A B = 3 , A D = 1 , D C = 2 , and B D A E = 2 5 , find the length of C F .
Also try Nasty Triangles 4 .
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Suppose B E = a , E C = b , A E = c and B D = 5 2 c , because A B = A C = 3 , ∠ A B C = ∠ A C B = ∠ E C D , so D E = E C = b . According to Stewart's theorem , A E 2 = B E + E C A B 2 × E C + A C 2 × B E − B E × E C c 2 = a + b 3 2 b + 3 2 a − a b = 9 − a b a b = 9 − c 2 According to secant-secant theorem , C D × C A = C E × C B b ( b + a ) = 2 × 3 a b + b 2 = 6 According to Ptolemy's theorem , A B × D E + A D × B E = A E × B D 3 b + a = 5 2 c 2 Now we have ⎩ ⎪ ⎨ ⎪ ⎧ a b = 9 − c 2 a b + b 2 = 6 3 b + a = 5 2 c 2 Solve the system of equations above and you would get a = 5 1 , b = 5 , c = 8 .
Now according to Menelaus' theorem , D A C D × F B A F × E C B E = 1 1 2 × B F 3 + B F × 5 5 1 = 1 So, B F = 2 , according to Menelaus' theorem again, B F A B × E D F E × C A D C = 1 2 3 × 5 F E × 3 2 = 1 Hence, F E = 5 , notice F E = E D = E C = 5 and F E D is a straight line, we can conclude ∠ F C D = 9 0 ∘ .
Thus, C F = A F 2 − A C 2 = 5 2 − 3 2 = 2 5 − 9 = 1 6 = 4
isn't there any method other than this i mean more simpler and little bit small steps?
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∠ A B C = ∠ A C B , ∠ A D E = 1 8 0 − ∠ A B C ⇒ ∠ C D E = ∠ A B C ⇒ △ A B C and △ D C E are similar ⇒ D E = E C , B C = B E + D E and B C 3 = 2 D E ⇒ D E × B C = 6 A B E D is cyclic ⇒ D B A E = A B × B E + A D × D E A B × A D + B E × D E = 2 5 From these we get D E = 5 , B E = 5 1 Using Menelaus Theorem: A F − 3 A F 5 1 1 2 = 1 we get A F = 5 Using Menelaus Theorem again: 2 3 E F D E 3 2 = 1 we get D E = E F = 5 = E C . From this we get ∠ A C F = 9 0 . C F = 5 2 − 3 2 = 4