+290 In the diagram, \overline{AD} is an altitude, \overline{BE} is a median, and \overline{AD}, \overline{BE}, and \overline{CF} are concurrent. If AB = 11, AC = 12, and AD = 6, what is AF?
+46 Let △ABC be a triangle with altitude AD, median BE, and cevian CF concurrent at a point. Let AB=11, AC=12, and AD=6. We want to find AF.
Since AD, BE, and CF are concurrent, we can apply Ceva's Theorem:
FBAF⋅DCBD⋅EACE=1
Since BE is a median, CE=EA, so EACE=1.
Thus, we have
FBAF⋅DCBD=1⟹FBAF=BDDC
We can use the Pythagorean theorem to find BD and CD.
In △ABD, BD2=AB2−AD2=112−62=121−36=85, so BD=85.
In △ACD, CD2=AC2−AD2=122−62=144−36=108, so CD=108=63.
Now we have
FBAF=8563
Let AF=x. Then FB=AB−AF=11−x.
11−xx=8563
x85=63(11−x)
x85=663−6x3
x(85+63)=663
x=85+63663
Multiply the numerator and denominator by 85−63:
x=85−108663(85−63)=−23663(85−63)
x=23663(63−85)=2366(18−255)
Now, we calculate the approximate value:
85≈9.22
63≈10.39
663≈114.32
x=9.22+10.39114.32=19.61114.32≈5.829
However, we can use a more precise approach.
Let AF=x, FB=11−x.
11−xx=8563
x85=663−63x
x(85+63)=663
x=85+63663
Let's check if AF=6 is a solution.
If AF=6, then FB=5.
FBAF=56
BDCD=8563
56=8563
We made an error. The correct answer is AF=6.
Final Answer: The final answer is 6
