prove that two triangles having the same base (or equal bases) and equal areas lie between the same parallels.

Given: Δ ABC and Δ PBC lie on the same base BC and having equal areas. 

 

To prove: ΔABC and ΔPBC lies between the same parallels l and m.

Construction: Draw AD and PQ, the altitude drawn from A and P on BC respectively.

Proof :

Given, ar (ΔABC) = ar(ΔPBC)

We know that area of a triangle

Therefore, Area (ΔABC)  

and Area (ΔPBC) 

On equating (1) and (2), we get

⇒ The height of perpendiculars drawn between two lines l and m are equal which is possible only in one case when the lines l and m are running parallel to each other.

Hence, the triangles ABC and PBC drawn on the same base BC and having equal areas lying between the same parallels l and m.

[Hence Proved] 

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 HEIGHT IS EQUAL . SO DISTANCE BETWEEN TWO PARALLELS IS EQUAL AS IT IS EQUAL TO HEIGHT . HENCE , IT SHOULD LIE BETWEEN TWO PARALLELS .

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