AA' BB' CC' are straight lines drawn from the angular points of a triangle through any point O within the triangle, and cutting the opposite sides at A', B', C'.
AP, BQ, CR are cut off from AA', BB', CC' and are equal to OA', OB',OC'.
Prove the area of triangle A'B'C' equals that of triangle PQR.
A rough diagram suggests that it might be that triangle OA'B' equals triangle OPQ, and similar for the other two . But with no restriction on the variables this will easily be shown by measurement not to be generally true.
I have no idea how to start. The source geometry book has, by this stage,
area = s(s-a)(s-b)(s-c)
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steve
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You probably mean heron's formula, which is $A=\sqrt{s(s-a)(s-b)(s-c)}$ with $s=\frac{a+b+c}{2}$. – Peter Jan 12 '15 at 20:09
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It is not true that $[OA'B']=[OPQ]$. – Jack D'Aurizio Jan 12 '15 at 20:10
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Let $x,y,z$ be the ratios $\frac{AC'}{C'B},\frac{BA'}{A'C},\frac{CB'}{B'A}$.
Due to Ceva's theorem, the lines $AA',BB',CC'$ concur in $O$ iff $xyz=1$. On the other hand, $$ [A'B'C']=[ABC]-[AB'C']-[BA'C']-[CA'B']=[ABC]\left(1-\frac{x}{z}-\frac{y}{x}-\frac{z}{y}\right)$$ so: $$ [A'B'C']=[ABC]\left(1-x^2 y-y^2 z-z^2 x\right).$$ Due to Van Obel's theorem, $\frac{AO}{OA'}=x+\frac{1}{z}$ and cyclic, so $\frac{OP}{OA}=1-\frac{1}{x+\frac{1}{z}}$ and $\frac{OQ}{OB}=1-\frac{1}{y+\frac{1}{x}}$, leading to: $$[OPQ]=\left(1-\frac{1}{x+\frac{1}{z}}\right)\left(1-\frac{1}{y+\frac{1}{x}}\right)[OAC]=\left(1-\frac{1}{x+\frac{1}{z}}\right)\left(1-\frac{1}{y+\frac{1}{x}}\right)\frac{OC'}{OC}[ABC]\\=\left(1-\frac{1}{x+\frac{1}{z}}\right)\left(1-\frac{1}{y+\frac{1}{x}}\right)\frac{1}{z+\frac{1}{y}}[ABC].$$ By adding these identities we easily get $[A'B'C']=[PQR]$ as wanted.
Jack D'Aurizio
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thank you for your solution. I will try to understand it. But as the question came from a secondary school book Plane Geometry Practical and Theoretical Books I II III (1903) (closely Euclid Books I II III) there must be a simpler answer ? – steve Jan 14 '15 at 19:38