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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 ]

922101_969202_ans_caf7a7169e7f40ac8da9485b668a7ebf.png

Mathematics

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