Question

Basic proportion theorem?
Std 10th

Answer 1

Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.

$<b >$ Proof :-

AD/DB = AE/EC

Construction: Join BE and CD Draw DM ⊥ AC and EN ⊥ AB.

Proof: Now, Now, ar (ADE) = 1/2 × Base × Height
= 1/2 × AE × DM ar (DEC)
= 1/2 × Base × Height
= 1/2 × EC × DM

Divide (3) and (4)
"ar (ADE)" /"ar (DEC)"
= (1/2 " × AE × DM" )/(1/2 " × EC × DM " )

"ar (ADE)" /"ar (DEC)"
= "AE" /"EC"

Now, ∆BDE and ∆DEC are on the same base DE and between the same parallel lines BC and DE.
∴ ar (BDE) = ar (DEC)
Hence, "ar (ADE)" /"ar (BDE)"
= "ar (ADE)" /"ar (DEC)" "AD" /"DB"
= "AE" /"EC"

Hence Proved!

Answer 2

Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.

 Proof :-

AD/DB = AE/EC

Construction: Join BE and CD Draw DM ⊥ AC and EN ⊥ AB. 

Proof: Now, Now, ar (ADE) = 1/2 × Base × Height 
= 1/2 × AE × DM ar (DEC) 
= 1/2 × Base × Height 
= 1/2 × EC × DM 

Divide (3) and (4) 
"ar (ADE)" /"ar (DEC)" 
= (1/2 " × AE × DM" )/(1/2 " × EC × DM " ) 

"ar (ADE)" /"ar (DEC)" 
= "AE" /"EC" 

Now, ∆BDE and ∆DEC are on the same base DE and between the same parallel lines BC and DE. 
∴ ar (BDE) = ar (DEC) 
Hence, "ar (ADE)" /"ar (BDE)" 
= "ar (ADE)" /"ar (DEC)" "AD" /"DB" 
= "AE" /"EC" 

Hence Proved!


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