In triangle $ABC$ , $X$ and $Y$ are points on sides $AC$ and $BC$ respectively . If $Z$ is on the segment $XY$ such that $\frac{AX}{XC} = \frac{CY}{YB} = \frac{XZ}{ZY}$ , then how to prove that the area of triangle $ABC$ is given by $[ABC]=([AXZ] ^{1/3} + [BYZ]^{1/3})^3.$

  • $\begingroup$ Thanks this give me the exact answer..... $\endgroup$ – OMJAISWAL Jun 18 '14 at 12:27


Your situation corresponds to De Casteljau's algorithm for quadratic Bézier curves. So let's call that common ratio

\begin{align*} \frac{t}{1-t} &= \frac{AX}{XC} = \frac{CY}{YB} = \frac{XZ}{ZY} \\ t &= \frac{AX}{AC} = \frac{CY}{CB} = \frac{XZ}{XY} \end{align*}

then you have

\begin{align*} X &= (1-t)A + tC \\ Y &= (1-t)C + tB \\ Z &= (1-t)X + tY = (1-t)^2A + 2t(1-t)C + t^2B \\ [AXZ] &= [A,(1-t)A+tC,(1-t)^2A + 2t(1-t)C + t^2B] \\&= [A,tC,t^2B] = t^3[ACB] \\ [BYZ] &= [B,(1-t)C+tB,(1-t)^2A + 2t(1-t)C + t^2B] \\&= [B,(1-t)C,(1-t)A] = (1-t)^3[BCA] \\ [AZX]^{1/3} &= t\sqrt[3]{[ABC]} \\ [BYZ]^{1/3} &= (1-t)\sqrt[3]{[ABC]} \\ [AZX]^{1/3} + [BYZ]^{1/3} &= (t+1-t)\sqrt[3]{[ABC]} = \sqrt[3]{[ABC]} \\ \left([AZX]^{1/3} + [BYZ]^{1/3}\right)^3 &= [ABC] \end{align*}

Note that I changed the orientation from $[AXZ]$ to $[AZX]$ to get the correct sign there. As long as you only care for the absolute value of the result, this should not bother you, although the computation relies on the use of negative areas for linear combinations.

I'm not sure how you interpret a notation like $[ABC]$. To me, the brackets signify a determinant of homogeneous coordinates:


Up to a scaling factor of $\frac12$ this determinant is equal to the triangle area (at least in the above case, where all columns have their last coordinate equal to $1$). If a column of the determinant is a sum of two vectors, you can split it into two determinants, one for each vector. If a column is scaled by some factor, you can move that factor in front of the determinant. If two columns are equal, the resulting determinant is zero. I've used all these rules in the computation above to move the polynomials in $t$ in front of the area computation.

You can also consider this in terms of areas, without resorting to determinants. For example, to show $[AZX]=t^3[ABC]$, start with $\triangle ABC$. You have $[AYC]=t[ABC]$ since you divide the edge $CB$ at position $t$. Likewise you have $[AYX]=t[AYC]$ as you divide $AC$. Last but not least, $[AZX]=t[AYX]$ as you divide $AX$. So on the whole, $[AZX]=t^3[ABC]$. Do the same for the other triangle, and you are done.

  • 1
    $\begingroup$ I think you flipped one of the ratios from OP's statement... $\endgroup$ – Hao Ye Jun 17 '14 at 5:33
  • $\begingroup$ @HaoYe: Thanks for catching my mistake! I completely rewrote my answer, and now can support the claim. $\endgroup$ – MvG Jun 17 '14 at 13:28

We can say that points $X$, $Y$, $Z$ separate respective segments $AC$, $CB$, $XY$ into sub-segments with ratios $b:r$ (think "blue : red"), where $b+r=1$.

enter image description here

Let's add a few perpendiculars, transfer our ratio-colors accordingly (as the reader may verify, using similar triangles), and compute the area of $\triangle AXZ$:

enter image description here

$$\begin{align} |\triangle AXZ| &= \frac{1}{2}\;|\overline{AX}|\;|\overline{RZ}| \\[6pt] &= \frac{1}{2}\cdot b\;|\overline{AC}|\cdot b\;|\overline{RS}| \\[6pt] &= \frac{1}{2}\cdot b\;|\overline{AC}|\cdot b\cdot b\;|\overline{PB}| \\[6pt] &= b^3\cdot \frac{1}{2}|\overline{AC}|\;|\overline{PB}| \\[6pt] &= b^3\;|\triangle ABC| \\[8pt] \text{Similarly, } |\triangle BYZ| &= r^3\;|\triangle ABC| \end{align}$$

Therefore, $$\frac{|\triangle AXZ|}{|\triangle ABC|} = b^3 \qquad\text{and}\qquad \frac{|\triangle BYZ|}{|\triangle ABC|} = r^3$$

Recalling $b+r=1$, the result follows. $\square$


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