prove that if a straight line is drawn parallel to any one side of a traingle, then it divides other two sides in equal ratio
Answers
Step-by-step explanation:
Theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points , then the other two sides are divided in the same ratio.
•Correct Question:-
If a straight line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
•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.