a) Trong hình có ba tam giác vuông là $\Delta $ AEB, $\Delta $ EBD, $\Delta $ BCD.

b) Ta có theo định lí Py-ta-go: $BE^{2}$ = $AE^{2}$ + $AB^{2}$ = $5^{2}$ + $7,5^{2}$ $\Rightarrow $ BE = 9cm

$\Delta $ AEB $\sim $ $\Delta $ CBD, ta có:

$\frac{AE}{BC}$ = $\frac{AB}{CD}$ $\Leftrightarrow $ $\frac{5}{6}$ = $\frac{7,5}{CD}$ $\Leftrightarrow $ CD = 9cm.

$\frac{AE}{BC}$ = $\frac{EB}{BD}$ $\Leftrightarrow $ $\frac{5}{6}$ = $\frac{9}{BD}$ $\Leftrightarrow $ BD = 10,8cm.

$ED^{2}$ = $BE^{2}$ + $BD^{2}$ = $9^{2}$ + $10,8{2}$ $\Rightarrow $ ED = 14,1cm.

c) S$\Delta $BDE = $\frac{1}{2}$.BE.BD

    S$\Delta $AEB = $\frac{1}{2}$.AE.AB

    S$\Delta $BCD = $\frac{1}{2}$.BC.CD

$\Rightarrow $ $\frac{S\Delta BDE}{S\Delta AEB + S\Delta BCD}$ = $\frac{BE.BD}{AE.AB + BC.CD}$ = $\frac{5.7,5}{6.9}$ = 1,06