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

Solution

## In $$\triangle BCX$$ and $$\triangle DCY$$,$$\angle CBX=\angle CDY$$          [ Corresponding angles ]$$\angle CXB=\angle CYD$$          [ Corresponding angles ]$$\therefore$$  $$\triangle BCX\sim \triangle DCY$$  [ By AA similarity ]We know that corresponding sides of similar triangles are proportional.$$\therefore$$  $$\dfrac{BC}{DC}=\dfrac{BX}{DY}=\dfrac{CX}{CY}$$$$\Rightarrow$$  $$\dfrac{BX}{DY}=\dfrac{BC}{DC}$$$$\Rightarrow$$  $$\dfrac{BX}{DY}=\dfrac{2DC}{DC}$$     [ As D is the mid-point of BC ]$$\Rightarrow$$  $$\dfrac{BX}{DY}=\dfrac{2}{1}$$    ---  ( 2 )Similarly $$\triangle AEX\sim ADY$$       [ By AA similarity ]$$\therefore$$  $$\dfrac{AE}{AD}=\dfrac{EX}{DY}=\dfrac{AX}{AY}$$$$\Rightarrow$$  $$\dfrac{EX}{DY}=\dfrac{AE}{AD}$$$$\Rightarrow$$  $$\dfrac{EX}{DY}=\dfrac{AE}{2AE}$$    [ As D is mid-point of BC ]$$\Rightarrow$$  $$\dfrac{EX}{DY}=\dfrac{1}{2}$$      ---- ( 2 )Dividing ( 1 ) by ( 2 ), we get$$\dfrac{BX}{EX}=4$$$$\Rightarrow$$  $$BX=4EX$$$$\Rightarrow$$  $$BE+EX=4EX$$$$\Rightarrow$$  $$BE=3EX$$$$\therefore$$   $$BE:EX=3:1$$                [ Proved ]Mathematics

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