Question

Prove that two triangles having the same base and equal areas lie between the same parallels.​

Answer 1

Answer:

Given:△ABC and △ABD are two trianles on same base AB such that area(△ABC)=area(△ABD).

Join CD

To prove:CD∥AB

Construction:Draw the altitutdes of △ABC and △ABD throught C and D and meeting base at E and F respectively.

Proof:Since CE⊥AB and DF⊥AB(by construction)

Since, lines perpendicular to same line are parallel to each other.

∴CE∥DF .......(1)

Now, area(△ABC)=

$\frac{1}{2}$

×AB×CE

area(△ABD)=

$\frac{1}{2}$

×AB×DF

Since area(△ABC)=area(△ABD)(Given)

∴CF=DF .........(2)

Now, In CDFE,CF∥DF and CE∥DF

Since one pair of opposite sides are equal and parallel

Hence CDFE is a parallelogram.

So, CD∥EF (opposite sides of parallelogram are parallel)

∴CD∥AB

Hence, proved

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