Theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Let us consider a triangle ABC in which a line that is parallel to side BC intersects other two sides AB and AC at D and E respectively. Hence we need to prove (AD/ DB) = (AE/ EC)
To prove the theorem, we need to join BE and CD and draw DM perpendicular to AC. Along with that, we need to draw EN perpendicular to AB as well.
The area of the triangle ADE = (0.5* base * height) = 0.5 * AD * EN
Now, considering AD as the base,
The area of ADE = 0.5 * AD * EN
Along with that,
The area of BDE = 0.5 * DB * EN
Now, considering AE as the base,
The area of ADE = 0.5 * AE * DM
And, the area of DEC = 0.5 * EC * DM
Hence, (The area of ADE / The area of BDE) = ((0.5 * AD * EN) / (0.5 * DB * EN)) = (AD / DB) ------------ equation number (1)
On the other hand, (The area of ADE / The area of DEC) = ((0.5 * AE * DM) / (0.5 * EC * DM)) = (AE / EC) ------------ equation number (2)
Now, The area of ADE = The area of DEC ------------ equation number (3)
Now, it is evident that the triangle BDE and DEC are on the same base which is DE and between the same parallels which are BC and DE.
Hence, from the equation number 1, 2 and 3, we get (AD / DB) = (AE / EC)
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