(a) The diagonal BD of a parallelogram ABCD intersects the line segment AE at the point F, where E is any point on the side BC. Prove that DF × EF =…

In \(\displaystyle \Delta \mathrm{ADF}\) and \(\displaystyle \Delta \mathrm{EBF}\),

\[\begin{aligned} & \angle \mathrm{DFA}=\angle \mathrm{EFB} \\ & \angle \mathrm{ADF}=\angle \mathrm{FBE} \\ & \therefore \triangle \mathrm{ADF} \sim \triangle \mathrm{EBF} \\ & \therefore \frac{\mathrm{DF}}{\mathrm{FB}}=\frac{\mathrm{FA}}{\mathrm{EF}} \\ & \Rightarrow \mathrm{DF} \times \mathrm{EF}=\mathrm{FB} \times \mathrm{FA} \end{aligned} \]

\[\begin{aligned} & \mathrm{AD}^{2}=\mathrm{BD} \times \mathrm{DC} \\ & \frac{\mathrm{AD}}{\mathrm{DC}}=\frac{\mathrm{BD}}{\mathrm{AD}} \end{aligned} \]

Also, \(\displaystyle \angle \mathrm{ADB}=\angle \mathrm{ADC}\)

\[\begin{aligned} & \therefore \Delta \mathrm{DBA} \sim \Delta \mathrm{DAC} \\ & \angle \mathrm{DBA}=\angle \mathrm{DAC} \\ & \angle \mathrm{BAD}=\angle \mathrm{DCA} \end{aligned} \]

Adding both

\[\begin{aligned} & \angle \mathrm{DBA}+\angle \mathrm{DCA}=\angle \mathrm{DAC}+\angle \mathrm{BAD} \\ & \therefore \angle \mathrm{BAC}=90^{\circ} \end{aligned} \]