Proving Area of Parallelogram PQRS with Geometry Proof

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To prove that the area of triangles TSR and TQP combined equals half the area of parallelogram PQRS, one must first establish the area formulas for both the parallelogram and the triangles. By relating these formulas, the common elements can be identified, which aids in the proof. An alternative approach involves comparing the areas of triangles PQT and PTS, as well as QTR and RTS, noting that appropriate choices of base and height yield equal areas. This method may simplify the computation required for the proof. Understanding these relationships is crucial for successfully demonstrating the area equivalence.
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PQRS is parallelogram and T is any point inside the parallelogram. Prove that delta TSR + delta TQP = 1/2 the area of parallelogram PQRS.

I know this problem has been posted earlier but there was no strong response.

Please someone help me out Grade 12 Geometry Mathematics is tough. :frown:
 
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Assuming that delta in your post means area, so to prove that:
A_{TPQ} + A_{TRS} = \frac{1}{2} A_{PQRS}, you need the formula to find the area for the parallelogram PQRS, the formala to find the area for the two triangles TPS, and TRS.
After that, you should relate the two formulae above. What do they have in common.
Ok, try the problem again, and see if you get it. :)
 
Another way to do it is to compare the area of PQT to the area of PTS, and by extension compare the areas of QTR and RTS. This might require a little less computation--you just have to observe that if you measure the triangles one way (choose the base and height appropriately) they must be equal.
 
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