Ceva's Theorem/Problems: Difference between revisions

Revision as of 15:16, 9 May 2021

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-==Introductory==
+#REDIRECT[[Ceva's theorem/Problems]]
-===I1===
-====Problem====
-Suppose <math>AB, AC</math>, and <math>BC</math> have lengths <math>13, 14</math>, and <math>15</math>, respectively.  If <math>\frac{AF}{FB} = \frac{2}{5}</math> and <math>\frac{CE}{EA} = \frac{5}{8}</math>,  find <math>BD</math> and <math>DC</math>.
-====Solution====
-If <math>BD = x</math> and <math>DC = y</math>, then <math>10x = 40y</math>, and <math>{x + y = 15}</math>.  From this, we find <math>x = 12</math> and <math>y = 3</math>.
-''[[Ceva's_Theorem | Back to main article]]''