Question

In a triangle ABC, E is point on side AC such that AE=4cm and EC=4cm. BC=7cm. D is a point on

AB such that AD=2cm. Find the length of BD.​

Answer 1

✬ BD = 4 cm ✬

Step by step explanation:

Given:

• E and D are two points on sides of ∆ABC.
• E is on side AC.
• D is on side AB.

To Find:

• Length of BD ?

Solution: Here we have

• AE = 4 cm
• EC = 4 cm
• BC = 7 cm
• AD = 2 cm

★ Thales ' theorem ★

• If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio.

In ∆ABC , we have

• AD/DB = AE/EC

$\implies{\rm }$ 2/DB = 4/4

$\implies{\rm }$ 2 × 4 = 4 × DB

$\implies{\rm }$ 8 = 4 × DB

$\implies{\rm }$ 8/4 = DB

$\implies{\rm }$ 2 = DB

Hence, the length of BD is 2 cm.

[ Verification ]

➮ AD/DB = AE/EC

➮ 2/2 = 4/4

➮ 1 = 1

$\large\bold{\texttt {Verified }}$

Answer 2

Answer:

✳️ Given ✳️

$\mapsto$ On a ∆ABC, E is the point on side AC such that AE = 4 cm and EC = 4 cm, BC = 7cm. D is a point on AB such that AD = 2 cm.

✳️ To Find ✳️

$\mapsto$ Find the length of BD.

✳️ Solution ✳️

Given :-

• AE = 4 cm
• EC = 4 cm
• BC = 7 cm
• AD = 2 cm

➕ By using Thales Theorem, [ As DE || BC ]

$\leadsto$ $\dfrac{AD}{BD}$ = $\dfrac{AE}{CE}$

• Let DB = x

$\implies$ $\dfrac{2}{x}$ = $\dfrac{4}{4}$

✍️ By doing cross multiplication we get,

:$\implies$ 4x = 8

:$\implies$ x = $\sf\dfrac{\cancel{8}}{\cancel{4}}$

:$\dashrightarrow$ x = 2

$\therefore$ The length of BD = 2.