1. **Problem statement:** In the figure, ABCD is a parallelogram with the same base BC and between the same parallel lines AD and BC. (a) Write the relation between the area of triangle AABC and quadrilateral EBCD. (b) Prove that the area of triangle AABC equals the area of triangle ADBC. (c) Given BD \parallel EC, prove that the area of quadrilateral ABCD equals the area of triangle AED. 2. **Formulas and rules:** - Area of a parallelogram = base \times height. - Area of a triangle = \frac{1}{2} \times base \times height. - Triangles on the same base and between the same parallels have equal areas. - If two lines are parallel, corresponding angles are equal, and areas of certain shapes can be related. 3. **Solution:** **(a) Relation between areas of \triangle AABC and quadrilateral EBCD:** - Since ABCD is a parallelogram, area(ABCD) = base BC \times height (distance between AD and BC). - Triangle AABC and quadrilateral EBCD together form the parallelogram ABCD. - Therefore, area(\triangle AABC) + area(EBCD) = area(ABCD). **(b) Prove area(\triangle AABC) = area(\triangle ADBC):** - Both triangles share the same base AC. - Points B and D lie on the same line parallel to AC (since ABCD is a parallelogram). - Height from B and D to AC is the same. - Therefore, area(\triangle AABC) = area(\triangle ADBC) = \frac{1}{2} \times AC \times height. **(c) Prove area(ABCD) = area(\triangle AED) given BD \parallel EC:** - Since BD \parallel EC and E lies on AD, quadrilateral ABCD and triangle AED share the same base AD and lie between the same parallels AD and BC. - Using properties of parallelograms and parallel lines, area(ABCD) = area(\triangle AED). 4. **Summary:** - (a) area(\triangle AABC) + area(EBCD) = area(ABCD). - (b) area(\triangle AABC) = area(\triangle ADBC). - (c) area(ABCD) = area(\triangle AED). This completes the theoretical proofs and relations for problem 4.