Question

if a line is drawn parallel to one side of the triangle to intersect the other two sides in distinct points , the other two sides are divided in the same ratio.​

Answer 1

Step-by-step explanation:

Given :-ABC is triangle where a line intersects sides AB and AC at points D and E respectively and DE II BC. To prove :-AD/DB=AE/EC construction :-Join DC and BE and draw DM perpendicular to side AE and Draw EN perpendicular to the side AD respectively. Proof:-Area of ADE=1/2* base*height =>1/2*AD*EN Area of BDE=1/2*base*height =>1/2*DB*EN Ar(ADE)/Ar(BDE)=1/2*AD*EN/1/2*DB*EN=AD/DB----------(i) Now Area of AED=1/2*base*height =>1/2*AE*DM Area of CED=1/2*base*height =>1/2*EC*DM Ar(AED)/Ar(CED)=1/2*AE*DM/1/2*EC*DM =AE/AC ---------------(ii) now in trapezium DEBC :-DE II BC DBE and CED lie on the same base DE and between the same parallels DE and BC respectively. So, Ar(DBE)=Ar(CED) - - - - - -(iii) So, from (I), (ii) and (iii) we obtain that, AD/DB=AE/EC (proved) .[If you like then please mark it as brainliest answer]