Question

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

Answer 1

Given :-

ΔABC where D is the mid point of BC and E is the midpoint of AC and BE produced cuts AC at X.

To prove :-

BE : EX = 3 : 1

Construction :-

Draw a line DY ║ BX and meeting AC at Y.

Proof :-

In ΔDCY and ΔBCX ,

∠c = ∠c  ( common )

∠BXC = ∠DYC ( Corresponding angles )

Therefore ΔDCY ~ ΔBCX ( AA Similarity )

Now,

( Corresponding sides are proportional )

= $\frac{DX}{BY} = \frac{BC}{DC} =\frac{CX}{CY}$

( As D is the midpoint of BC )

= $\frac{BX}{DY} = \frac{2CD}{CD}$

So,

$\frac{BX}{DY} =2 -----1$

In ΔAEX and ΔADY ,

( As E is the midpoint of AC )

= $\frac{EX}{DY} =\frac{AE}{ED} =\frac{1}{2} -------2$

Now dividing 1 and 2 ,

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

= BE + EX = 4EX

= BE = 3EX

BE : EX = 3 : 1 Proved.

Answer 2

Solution:

Given that: In ΔABC, D is the midpoint of side BC. AD is bisected at point E.

To prove - BE : EX = 3 : 1

$\tt \implies \dfrac{BE}{EX}= \dfrac{3}{1}$

Construction: Take a point Y at AC and join to D such that BX || DY.

Proof: In ΔBXC and ΔDYC

⇒ ∠C = ∠C           [common]

⇒ ∠XBC = ∠YDC    [Because, BX || DY]

∴ ΔBXC ~ ΔDYC      [AA similarity criterion]

Now, we know that the sides of similar triangles are proportional to each other.

$\implies \tt \dfrac{BX}{DY}= \dfrac{BC}{DC}$

But, D is the midpoint of side BC.

⇒ BC = 2DC

$\implies \tt \dfrac{BX}{DY}= \dfrac{2DC}{DC}$

$\implies \tt \dfrac{BX}{DY}= 2 \: \sf .... (i)$

$\rule{300}{2}$

In ΔAEX and ΔADY

⇒ ∠A = ∠A      [common]

⇒ ∠AEX = ∠ADY    [Because, EX || DY]

∴ ΔAEX ~ΔADY      [By AA criterion]

$\implies \tt \dfrac{EX}{DY}= \dfrac{AE}{AD}$

But, E is the midpoint of AD,

⇒ AD = 2AE

$\implies \tt \dfrac{EX}{DY}= \dfrac{AE}{2AE}$

$\implies \tt \dfrac{EX}{DY}= \dfrac{1}{2} \: \: \sf ...(ii)$

On dividing (i) and (ii),

$\implies \tt \dfrac{BX}{DY} \div \dfrac{EX}{DY} = 2 \div \dfrac{1}{2}$

$\implies \tt \dfrac{BX}{EX}= 4$

⇒ BX = 4 EX

⇒ BX - EX = 4 EX

⇒ 4EX - EX = BX

⇒ 3EX = BX

$\tt \implies \dfrac{BE}{EX}= \dfrac{3}{1}$

$\therefore$ BX : EX = 3 : 1

            Hence, it is proved.