Question

prove that if a straight line is drawn parallel to a side of triangle then it​

Answer 1

•Given :

In a triangle ABC, a straight line l parallel to BC, intersects AB at D and AC at E.  

•To prove :

AD/DB  =  AE/EC

•Construction :

Join BE, CD.  

Draw EF ⊥ AB and DG ⊥ CA

•Proof :

Step:1

Because EF ⊥ AB, EF is the height of the triangles ADE and DBE.  

Area (ΔADE)  =  1/2 ⋅ base ⋅ height  =  1/2 ⋅ AD ⋅ EF

Area (ΔDBE)  =  1/2 ⋅ base ⋅ height  =  1/2 ⋅ DB ⋅ EF

Therefore,  

Area (ΔADE) / Area (ΔDBE)  :

=  (1/2 ⋅ AD ⋅ EF) / (1/2 ⋅ DB ⋅ EF)

Area (ΔADE) / Area (ΔDBE)  =  AD / DB -----(1)

Step:2  

Similarly, we get

Area (ΔADE) / Area (ΔDCE)  :

=  (1/2 ⋅ AE ⋅ DG) / (1/2 ⋅ EC ⋅ DG)

Area (ΔADE) / Area (ΔDCE)  =  AE / EC -----(2)

Step:3

But ΔDBE and ΔDCE are on the same base DE and between the same parallel straight lines BC and DE.  

Therefore,  

Area (ΔDBE)  =  Area (ΔDCE) -----(3)

Step:4

From (1), (2) and (3), we can obtain

AD / DB  =  AE / EC

Hence, the theorem is proved.