In the following figure `AP_|_BC, AD||BC,` then find `A(DeltaABC):A(De

1.	In the following figure `AP_\|_BC, AD\|\|BC,` then find `A(DeltaABC):A(DeltaBCD)`.
Answer» `DeltaABC` and `DeltaBCD` have same base BC. The vertices of `DeltaABC` and `DeltaBCD` lie on the parallel lines AD and BC. The distance between the two parallel lines is the height of the triangles. Distance between parallel lines is constant. `:.DeltaABC` and `DeltaBCD` have equal heights. Areas of two traingles of equal bases and equal heights are equal. `:.A(DeltaABC)=A(DeltaBCD)` `:.(A(DeltaABC))/(A(DeltaBCD))=1/1` `:.A(DeltaABC):A(DeltaBCD)=1:1` `A(DletaABC):A(DeltaBCD)=1:1`

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In the following figure `AP_|_BC, AD||BC,` then find `A(DeltaABC):A(DeltaBCD)`.

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