Question

if in ABC,a line parllel to BC intersect other two sides AB & AC at D & E respectively, then prove that AD/ AB = AE/AC​

Answer 1

Step-by-step explanation:

Given:

• In ∆ABC ,
• DE is parallel to BC.
• DE intersects AB at D and AC at E.

To Prove:

• AD/AB = AE/AC

Proof: In ∆ABC , DE is parallel to BC so, by Thales' Theorem we have,

$\implies{\rm }$ AD/DB = AE/EC or,

$\implies{\rm }$ DB/AD = EC/AE

Now, Adding 1 to both LHS & RHS

$\implies{\rm }$ DB/AD + 1 = EC/AE + 1 (Take LCM)

$\implies{\rm }$ DB + AD/AD = EC + AE/AE

$\implies{\rm }$ AB/AD = AC/AE or,

$\implies{\rm }$ AD/AB = AE/AC

Hence, it is proved.

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Thales' theorem or Basic Proportionality Theorem (BPT) states that if a line is drawn parallel to one side of a triangle to the intersect the other sides in distinct points then the two sides are divided in the same ratio.