Question

D is the midpoint of side BC of triangle ABC and Ad is Bisected
at the point E and BE produce d to cuts AC at point x prove that
BE:EX=3:1
correct ans will perfectly mark as brainliest and be follpwed​

Answer 1

$\huge\sf\blue{Given}$

➢ D is the midpoint of side BC of ∆ABC

➢ AD is the bisected at point E

➢ BE is produced to D and cuts AC at X

$\rule{110}1$

$\huge\sf\gray{To\;Prove}$

✭ BE : EX = 3:1

$\rule{110}1$

$\huge\sf\purple{Steps}$

To prove - BE : EX = 3 : 1

$\sf\mapsto\dfrac{BE}{EX}= \dfrac{3}{1}$

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.

$\leadsto \sf\dfrac{BX}{DY}= \dfrac{BC}{DC}$

But, D is the midpoint of side BC.

⇒ BC = 2DC

$\leadsto \sf\dfrac{BX}{DY}= \dfrac{2DC}{DC}$

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

In ΔAEX and ΔADY

⇒ ∠A = ∠A [common]

⇒ ∠AEX = ∠ADY [Because, EX || DY]

∴ ΔAEX ~ΔADY [By AA criterion]

$\leadsto \sf\dfrac{EX}{DY}= \dfrac{AE}{AD}$

But, E is the midpoint of AD,

⇒ AD = 2AE

$\dashrightarrow \sf \dfrac{EX}{DY}= \dfrac{AE}{2AE}$

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

On dividing (i) and (ii),

➝ $\sf \dfrac{BX}{DY} \div \dfrac{EX}{DY} = 2 \div \dfrac{1}{2}$

➝ $\sf\dfrac{BX}{EX}= 4$

➝ BX = 4 EX

➝ BX - EX = 4 EX

➝ 4EX - EX = BX

➝ 3EX = BX

$\sf\color{aqua}{ \dfrac{BE}{EX}= \dfrac{3}{1} }$

Hence Proved!!

$\rule{170}3$