What is Thales Theorem??Explain.
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It's Thales theorem
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→Thales theorem is also known as Basic Proportionality 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 sided are divided in the same ratio.
Given,
In ∆ABC, DE || BC which intersects sides AB and AC at D and E respectively,
Required to prove;-
AD/DB = AE/EC
Construction:'
Join B,E and C,D and then draw DM ⊥ AC and EN ⊥ AB.
Proof:-
Area of ∆ADE = ½ × AD × EN
Area of ∆BDE = ½ × BD × EN
ar (∆ADE)/ar (∆BDE) = (½ × AD × EN)/(½ × BD × EN)
= AD/BD -------(1)
Area of ∆ADE = ½ × AE × DM
Area of ∆CDE = ½ × EC × DM
ar (∆ADE)/ar (∆CDE) = (½ × AE × DM)/(½ × EC × DM)
= AE/EC ------(2)
Observe that ∆BDE and ∆CDE are on the same base DE and between same parallels BC and DE,
ar (∆BDE) = ar (∆CDE) ------(3)
From (1) (2) and (3), we get
AD/DB = AE/EC
Hence proved.
Hope it helps
→ 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 sided are divided in the same ratio.
Given,
In ∆ABC, DE || BC which intersects sides AB and AC at D and E respectively,
Required to prove;-
AD/DB = AE/EC
Construction:'
Join B,E and C,D and then draw DM ⊥ AC and EN ⊥ AB.
Proof:-
Area of ∆ADE = ½ × AD × EN
Area of ∆BDE = ½ × BD × EN
ar (∆ADE)/ar (∆BDE) = (½ × AD × EN)/(½ × BD × EN)
= AD/BD -------(1)
Area of ∆ADE = ½ × AE × DM
Area of ∆CDE = ½ × EC × DM
ar (∆ADE)/ar (∆CDE) = (½ × AE × DM)/(½ × EC × DM)
= AE/EC ------(2)
Observe that ∆BDE and ∆CDE are on the same base DE and between same parallels BC and DE,
ar (∆BDE) = ar (∆CDE) ------(3)
From (1) (2) and (3), we get
AD/DB = AE/EC
Hence proved.
Hope it helps
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