Question

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

Answer 1

In $△BCX$ and $△DCY$,

$\angle CBX=\angle CDY$ [ Corresponding angles ]

$\angle CXB=\angle CYD$ [ Corresponding angles ]

$\therefore$ $△BCX\sim △DCY$ [ By AA similarity ]

We know that corresponding sides of similar triangles are proportional.

$\therefore$ $\frac{BC}{DC}=\frac{BX}{DY}=\frac{CX}{CY}$

$\Rightarrow$ $\frac{BX}{DY}=\frac{BC}{DC}$

$\Rightarrow$ $\frac{BX}{DY}=\frac{2DC}{DC}$ [ As D is the mid-point of BC ]

$\Rightarrow$ $\frac{BX}{DY}=\frac{2}{1}$ --- ( 2 )

Similarly $△AEX\sim ADY$ [ By AA similarity ]

$\therefore$ $\frac{AE}{AD}=\frac{EX}{DY}=\frac{AX}{AY}$

$\Rightarrow$ $\frac{EX}{DY}=\frac{AE}{AD}$

$\Rightarrow$ $\frac{EX}{DY}=\frac{AE}{2AE}$ [ As D is mid-point of BC ]

$\Rightarrow$ $\frac{EX}{DY}=\frac{1}{2}$ ---- ( 2 )

Dividing ( 1 ) by ( 2 ), we get

$\frac{BX}{EX}=4$

$\Rightarrow$ $BX=4EX$

$\Rightarrow$ $BE+EX=4EX$

$\Rightarrow$ $BE=3EX$

$\therefore$ $BE:EX=3:1$ [ Proved ]