Triangles on the same base and between the same parallels are equal in area. GIVEN : Two triangles `A B C` and `P B C` on the same base `B C` and between the same parallel lines `B C` and `A Pdot` TO PROVE : `a r( A B C)=a r( P B C)` CONSTRUCTION : Through `B ,` draw `B D C A` intersecting `P A` produced in `D` and through `C ,` draw `C Q B P ,` intersecting line `A P` in Q.

Explanation

Proof: `△ABC` & `△ABD` are drawn on the same base `AB` as base lying between the same two parallel lines `L_1` & `L_2`.
Here we will be using the following property of a triangle and parallelogram drawn on the same base between two parallel lines.
Since `△ABC` & parallelogram `ABPQ` are on the same base and between the same parallel `L_1` & `L_2`.
`∴ar(△ABC)=1/2ar(ABPQ)`
Similarly,
`ar(△ABD)=1/2ar(ABPQ)`
`∴ar(△ABC)=ar(△ABD)`
Hence Proved.