Let ABCD be an isosceles trapezium with parallel sides AB and CD, where AB>CD. P is a point inside ABCD such that the areas of triangles PCD, PBC, PBA and PAD are 3 $cm^2$, 4 $cm^2$, 5 $cm^2$, 6 $cm^2$ respectively. What is the ratio of AB:CD ?
My attempt :-
I can get total area in terms of A1 and A2 [where A1 = area of $\Delta$ AOB and A2= area of $\Delta$ COD as per figure]
$\sqrt{A1} + \sqrt{A2} = 3\sqrt{2}$
Also I know $(\frac{AB}{CD})^2$ = $\frac{A1}{A2}$ since $\Delta$ AOB $\sim$ $\Delta$ COD
How do I proceed further now ?
Apart from this I know that all isoscles trapezoids are cyclic quadrilaterals as well, I am not sure if it could be used here
Best Answer
Since the point $P$ may be anywhere within the trapezium, I'm unsure of how to use information with the diagonals, including their intersection point $O$. Also, I don't see any way to use that isosceles trapezoid are cyclic quadrilaterals. Instead, consider the diagram below,
where $EPG$ is parallel to $AB$ and $CD$ (with $E$ being on $AD$ and $G$ being on $BC$), $CFH$ is perpendicular to $AB$ (with $F$ being on $EG$ and $H$ being on $AB$), $\lvert CD\rvert = x$, $\lvert AB\rvert = rx$ (so the ratio being requested is $r$), $\lvert CF\rvert = h_1$, $\lvert FH\rvert = h_2$, $\lvert FG\rvert = z_1$ and $\lvert HB\rvert = z_2$.
Since $\triangle CFG \sim \triangle CHB$, we have
$$\frac{z_1}{h_1} = \frac{z_2}{h_1 + h_2} \;\;\to\;\; z_1(h_1+h_2) = z_2h_1 \tag{1}\label{eq1A}$$
Using the one-half base times height formula for triangle area with $\triangle PCD$ and $\triangle PAB$, we get
$$3 = \frac{xh_{1}}{2} \;\;\to\;\; xh_{1} = 6 \tag{2}\label{eq2A}$$
$$5 = \frac{(rx)h_{2}}{2} \;\;\to\;\; rxh_{2} = 10 \;\;\to\;\; xh_{2} = \frac{10}{r} \tag{3}\label{eq3A}$$
In addition, since $ABCD$ is an isosceles trapezium, with $AB \parallel CD$, we also have
$$rx = x + 2z_2 \;\;\to\;\; 2z_2 = (r-1)x \tag{4}\label{eq4A}$$
Splitting $\triangle PAD$ and $\triangle PBC$ into the upper & lower parts, then using the sum of areas of triangles on them,
$$6 = \frac{1}{2}\lvert EP\rvert(h_1+h_2) \;\;\to\;\; 12 = \lvert EP\rvert(h_1+h_2) \tag{5}\label{eq5A}$$
$$4 = \frac{1}{2}\lvert PG\rvert(h_1+h_2) \;\;\to\;\; 8 = \lvert PG\rvert(h_1+h_2) \tag{6}\label{eq6A}$$
Adding \eqref{eq5A} and \eqref{eq6A}, then using that $\lvert EP\rvert + \lvert PG\rvert = \lvert EG\lvert = x + 2z_1$, along with \eqref{eq1A}, \eqref{eq4A}, \eqref{eq2A} and \eqref{eq3A}, we get
$$\begin{equation}\begin{aligned} 20 & = (x + 2z_1)(h_1 + h_2) \\ & = x(h_1 + h_2) + 2(z_1(h_1 + h_2)) \\ & = xh_{1} + xh_{2} + 2(z_2h_1) \\ & = xh_{1} + xh_{2} + (r - 1)xh_1 \\ & = xh_{1} + xh_{2} + rxh_1 - xh_1 \\ & = r(xh_1) + xh_{2} \\ & = 6r + \frac{10}{r} \end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Multiplying both sides by $\frac{r}{2}$, and rearranging, gives
$$3r^2 - 10r + 5 = 0 \tag{8}\label{eq8A}$$
Using the quadratic formula, and that $r \gt 1$, we finally get
$$r = \frac{10 \pm \sqrt{100 - 60}}{6} = \frac{5 \pm \sqrt{10}}{3} \;\;\to\;\; r = \frac{5 + \sqrt{10}}{3} \approx 2.72 \tag{9}\label{eq9A}$$