Answer:
Given: A triangle ABC and a line DE parallel to BC.
To prove: A line parallel to one side of a triangle divides the other two sides proportionally.
Proof: Consider ΔABC and DE be the line parallel to Bc, then from ΔABC and ΔADE, we have
∠A=∠A (Common)
∠ADE=∠ABC (Corresponding angles)
Thus, by AA similarity, ΔABC is similar to ΔADE, therefore
AB/AD= AC/AE
⇒AD+DB/AD = AE+EC/AE
⇒1+DB/AD = 1+ EC/AE
⇒DB/AD = EC/AE
Therefore, a line parallel to one side of a triangle divides the other two sides proportionally.
⇒Therefore Proved
Hope this helps!!!
