Question

Please give the proof of the corollary derived from BPT that DE/BC = AE/AC=AD/AB where DE ll BC

Answer 1

since DE║BC and AB is transversal,then

∠ADE=   ∠ABC (corresponding angle)

Since  DE║BC and AC is transversal, then

    ∠AED=    ∠ACB [corresponding angle]

In Δ ADE and     ΔABC,
 
    ∠ADE=      ∠ABC [proved above]
    ∠AED=    ∠ ACB[proved above]

==>    ΔADE~   ΔABC

 =>    AD/AB= DE/BC= AE/AC [corresponding sides of similiar Δ's are proportional]
 
 
                                                                                               PROVED