For reference: In the figure, $ABCD$ is an isosceles trapezoid ( $BC \parallel AD$),
$M, N, P$ and $Q$ are midpoints and $S_{AMR} = 4 m^2$
Calculate: $S1.S2$ (Answer:$64m^2$)
I couldn't understand this resolution .. could anyone clarify or present another solution?
$MN = QP = \frac{AC}{2}\\ NP = MQ = \frac{BD}{2} = \frac{AC}{2}\\ BD \cap MN = X\\ \triangle BXM \cong \triangle AMR \\ XNTJ \sim TQRS \sim MNPQ \implies \boxed{S_{MXTR}=8 \implies \sqrt{S1.S2}=8}???\\ \therefore S1.S2 = 64$
Now, I was able to understand $S_{ABT} = 4S_{RMA} = 16\\ S_{BMX} = S_{RMA} = 4 \therefore S_{XMRT} = 8 = S_{TJPS}\\ S1.S2 =8^2 = 64$
$S_{MXTR} = 2 \cdot S_{\triangle ARM} =8$
Note that $NJTX$ and $RTSQ$ are rhombus and hence $S_{MXTR} = \sqrt{S_1 \cdot S_2}$. Here is another way to look at it -
By midpoint theorem, $MQ \parallel BD$ and $AR = RT$. Similarly, $MN \parallel AC$
Also, can you see $\triangle ARQ \cong \triangle DSQ$ and $\triangle BXN \cong \triangle CJN$?
So, $NJTX$ and $RTSQ$ are rhombus with lengths $RM$ and $AR$ respectively. If $\angle ARM = \theta$,
$S_1 = RM^2 \sin\theta, S_2 = AR^2 \sin\theta$
$S_1 \cdot S_2 = (2 \cdot \frac 1 2 \cdot AR \cdot RM \cdot \sin\theta)^2 = (2 \cdot S_{\triangle ARM})^2 = 64$