Question

In a trapezium ABCD, AB|| DC and DC = 2AB, FE drawn parallel to AB cuts AD in F and BC in E such that 4BE = 3EC. Diagonal DB intersects EF at G. Prove that 7FE = 10 AB. [4 MARKS]

Answer 1

Application of theorems: 2 Marks
Calculation: 2 Marks
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Proof: GE||DC[AB||DC and EF || AB]
$\begin{array}{ccc}\Rightarrow & \Delta bge\sim \Delta BDC& \left[AAsimilarly\right]\\ \therefore & \frac{BG}{BD}=\frac{BE}{BC}=\frac{GE}{DC}& \\ \Rightarrow & \frac{BG}{BD}=\frac{GE}{DC}=\frac{3}{7}& \{\because \frac{BE}{EC}=\frac{E}{4}\Rightarrow \frac{BE}{BE}=\frac{3}{7}\}\\ \Rightarrow & \frac{GE}{DC}=\frac{3}{7}and\frac{BG}{BD}=\frac{3}{7}& \\ \Rightarrow & GE=\frac{3}{7}DC& \\ \Rightarrow & GE=\frac{3}{7}\times 2AB=\frac{6}{7}AB& \\ \Rightarrow & GE=\frac{6}{7}AB& \dots \left(i\right)\\ \therefore & \frac{DG}{BC}=\frac{4}{7}& \\ & \angle GDF=\angle BDA& \left(Common\right)\\ & \angle DFG=\angle DAB& \left(Correspondingangle\right)\\ \therefore & \Delta DFG\sim \Delta DAB& \\ \therefore & \frac{FG}{AB}=\frac{DG}{BC}\Rightarrow \frac{FG}{AB}=\frac{4}{7}& \\ \Rightarrow & fg=\frac{4}{7}AB& \end{array}$
Adding (i) and (ii), we get
$\begin{array}{cc}\Rightarrow & GE+FG=\frac{6}{7}AB+\frac{4}{7}AB\\ \Rightarrow & FE=\frac{10}{7}AB\Rightarrow 7EF=10AB\end{array}$