Basic proportion theorem?
Std 10th
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Answered by
6
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!
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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rakesh4860:
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Answered by
3
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!
hope it helps u❤️❤️
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!
hope it helps u❤️❤️
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