Drawing an isosceles triangle with Cevians from the base

Question

I would like to draw an isosceles triangle with certain conditions. $\triangle{ABC}$ is an isosceles triangle, and $AB$ is its base. From the endpoints of the base, cevians $AD$ and $BE$ are drawn. They partition the region bound by $\triangle{ABC}$ into four regions. The area of $\triangle{ABD}$ is 10, the area $\triangle{ABE}$ is 9, the area of the region under both cevians is 6, and the area of the remaining region is $(10+9)/2 = 9.5$. $AE : CE$ is $2:3$, and $BD : CD$ is $4:5$.

Answer 1

\begin{align} S_{ABD}&=10 ,\quad S_{ABE}=9 ,\quad S_{ABP}=6 ,\quad S_{CEPD}=9.5 ,\\ S_{ABC}&=S_{ABD}+S_{ABE}-S_{ABP}+S_{CEPD} =\frac {45}2 ,\\ |AC|=|BC|&=a ,\\ |BD|&=\tfrac49a ,\\ |CD|&=\tfrac59a ,\\ |AE|&=\tfrac25a ,\\ |CE|&=\tfrac35a . \end{align}

Let $|CH|=h_c$. Then \begin{align} |AB|&=c=\frac{2S_{ABC}}{h_c}=\frac{45}{h_c} ,\\ |AC|=|BC|&=a=\sqrt{\frac{c^2}4+h_c^2} = \tfrac12\sqrt{\frac{2025}{h_c^2}+4h_c^2} . \end{align}

Using Stewart’s theorem, we can express $|AD|^2$ and $|BE|^2$ in terms of $h_c$:

\begin{align} c^2\cdot \tfrac59a+ a^2\cdot \tfrac49a&=a(|AD|^2+\tfrac59\cdot\tfrac49 a^2) ,\\ c^2\cdot \tfrac35a+ a^2\cdot \tfrac25a&=a(|BE|^2+\tfrac35\cdot\tfrac25 a^2) ,\\ |AD|^2&=\frac{99225+16h_c^4}{81h_c^2} .\\ |BE|^2&= \tfrac4{25}\cdot\frac{8100+h_c^4}{h_c^2} . \end{align}

Using a Heron’s formula for the area in terms of the squares of the sides,

\begin{align} S&=\tfrac14\sqrt{4a^2b^2-(a^2+b^2-c^2)^2} \end{align} for $\triangle ABD$, $\triangle ABE$ we can conclude that these areas does not depend on $h_c$, so the stated set of constraints is not enough for unique solution:

choose any $h_c$, set $c=\frac{45}{h_c}$ and the points $A,B,C$ would be defined. Then split $AC,BC$ as stated to get $D$ and $E$.

For example, in the picture one more constraint was used: $a=c$, to get equilateral triangle. In this case we get

\begin{align} h_c&=\sqrt{\frac{45\sqrt3}2}\approx 6.242687 ,\\ a=b=c&=3^{3/4}\sqrt{10} . \end{align}

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\begin{align} S_{ABD}&= \tfrac14\cdot \sqrt{4\cdot c^2\cdot|AD|^2-(c^2+|AD|^2-(\tfrac 49 a)^2)^2} \\ &= \tfrac14\cdot \sqrt{4\cdot \left(\frac {45}{h_c}\right)^2 \cdot \frac{99225+16h_c^4}{81h_c^2} - \left(\left(\frac {45}{h_c}\right)^2+ \frac{99225+16h_c^4}{81h_c^2} -\frac{16}{81} \left( \frac{2025}{4h_c^2}+h_c^2 \right) \right)^2} \\ &= \tfrac14\,\sqrt{\frac{9922500+1600h_c^4}{hc^4}-\frac{9922500}{h_c^4}} \equiv 10 . \end{align}

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Ups, in fact, it's much simpler:

\begin{align} S_{ABD}&= \frac{S_{ABD}}{S_{ABC}}\cdot S_{ABC} \\ &= \frac{\tfrac12\cdot \tfrac49 a\cdot h_a}{\tfrac12\cdot a\cdot h_a} \cdot S_{ABC} \\ &=\frac49\cdot S_{ABC}=\frac49\cdot \frac{45}2 \equiv 10 . \end{align}