Question

proof of basic proportionality theorem

Answer 1

u mean thales theorem?

Answer 2

Given: DE ll BC
T.P: AD/BD = AE/EC
Construction: Join BE and CD. Also draw EN perpendicular to AB and DM perpendicular to AC.
Proof: ar(ADE) = 1/2 * AD * EN
ar(BDE)= 1/2 * BD * EN
Taking the ratio,
1/2 * AD * EN/ 1/2 * BD * EN = AD/BD - (1)
Similarly,
ar(ADE)/ar(CDE) = 1/2 * AE * DM/ 1/2 * CE * DM = AE/EC - (2)
We have learnt earlier that if the two triangles have the same base and are between the same parallels, their areas are equal.
Here, BDE and CDE have the same base DE and are between the same parallels.
Therefore, ar(BDE) = ar(CDE) - (3)

From (1), (2) and (3),
AD/BD = AE/EC