Question

D is the mid-point of side BC of a ∆ABC. AD is bisected at the point E and BE produced cuts AC at the point X. Prove that BE = EX = 3 : 1

Answer 1

Given: ABC is a triangle in which D is the mid point of BC, E is the mid point of AD. BE produced meets AC at X.
To Prove: BE : EX = 3:1.
Construction: We draw a line DY parallel to BX.

Proof:

$$\begin{gathered} \mathrm{In}∆\mathrm{BCX}\mathrm{and}∆\mathrm{DCY}, \\ \angle \mathrm{CBX}=\angle \mathrm{CDY}\left(\mathrm{Corresponding}\mathrm{angles}\right) \\ \angle \mathrm{CXB}=\angle \mathrm{CYD}\left(\mathrm{Corresponding}\mathrm{angles}\right) \\ △\mathrm{BCX}~△\mathrm{DCY}\left(\mathrm{AA}\mathrm{similarity}\right) \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{corresponding}\mathrm{sides}\mathrm{of}\mathrm{similar}\mathrm{triangles}\mathrm{are}\mathrm{proportional}. \\ \mathrm{Thus},\frac{\mathrm{BC}}{\mathrm{DC}}=\frac{\mathrm{BX}}{\mathrm{DY}}=\frac{\mathrm{CX}}{\mathrm{CY}} \\ \Rightarrow \frac{\mathrm{BX}}{\mathrm{DY}}=\frac{\mathrm{BC}}{\mathrm{DC}} \\ \Rightarrow \frac{\mathrm{BX}}{\mathrm{DY}}=\frac{2\mathrm{DC}}{\mathrm{DC}}\left(\mathrm{As}\mathrm{D}\mathrm{is}\mathrm{the}\mathrm{mid}\mathrm{point}\mathrm{of}\mathrm{BC}\right) \\ \Rightarrow \frac{\mathrm{BX}}{\mathrm{DY}}=\frac{2}{1}....\left(1\right) \\ \mathrm{In}∆\mathrm{AEX}\mathrm{and}∆\mathrm{ADY}, \\ \angle \mathrm{AEX}=\angle \mathrm{ADY}\left(\mathrm{Corresponding}\mathrm{angles}\right) \\ \angle \mathrm{AXE}=\angle \mathrm{AYD}\left(\mathrm{Corresponding}\mathrm{angles}\right) \\ △\mathrm{AEX}~△\mathrm{ADY}\left(\mathrm{AA}\mathrm{similarity}\right) \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{corresponding}\mathrm{sides}\mathrm{of}\mathrm{similar}\mathrm{triangles}\mathrm{are}\mathrm{proportional}. \\ \mathrm{Thus},\frac{\mathrm{AE}}{\mathrm{AD}}=\frac{\mathrm{EX}}{\mathrm{DY}}=\frac{\mathrm{AX}}{\mathrm{AY}} \\ \Rightarrow \frac{\mathrm{EX}}{\mathrm{DY}}=\frac{\mathrm{AE}}{\mathrm{AD}} \\ \Rightarrow \frac{\mathrm{EX}}{\mathrm{DY}}=\frac{\mathrm{AE}}{2\mathrm{AE}}\left(\mathrm{As}\mathrm{D}\mathrm{is}\mathrm{the}\mathrm{mid}\mathrm{point}\mathrm{of}\mathrm{BC}\right) \\ \Rightarrow \frac{\mathrm{EX}}{\mathrm{DY}}=\frac{1}{2}....\left(2\right) \\ \mathrm{Dividing}\left(1\right)\mathrm{by}\left(2\right),\mathrm{we}\mathrm{get} \\ \frac{\mathrm{BX}}{\mathrm{EX}}=4 \\ \Rightarrow \mathrm{BX}=4\mathrm{EX} \\ \Rightarrow \mathrm{BE}+\mathrm{EX}=4\mathrm{EX} \\ \Rightarrow \mathrm{BE}=3\mathrm{EX} \\ \Rightarrow \mathrm{BE}:\mathrm{EX}=3\hspace{0.17em}:1 \end{gathered}$$