Prove Parallelogram ABCD: Triangle APD = ABP + DCP

AI Thread Summary
In parallelogram ABCD, the task is to prove that the area of triangle APD equals the sum of the areas of triangles ABP and DCP when point P is on BC. The key to the proof lies in recognizing that the bases of triangles ABP and DCP together equal the base of triangle APD. All three triangles share the same height, which allows for a straightforward area comparison. Understanding this relationship between the bases and heights is crucial for completing the proof. This geometric property highlights the additive nature of areas within the parallelogram.
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Parallelogram proof :(

Homework Statement


In parallelogram ABCD, P is any point on BC. Prove that triangle APD = triangle ABP + triangle DCP

Homework Equations


n/a

The Attempt at a Solution


I really don't know where to start :(

any help would be greatly appreciated :)
 
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First do you mean that the areas satisfy that equation? If so just note that the lengths of the bases of the two triangles ABP and DCP add to the length of the base of triangle APD and all three triangles have the same height.
 

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