(i) In D A C, S is the mid point of DA and R is the mid point of DC. So |AO|=|OC|, and we have our second side for the Side-Angle-Side postulate.Īnd if triangles ΔAOD and ΔCOD are congruent, |AD|=|DC|, and we are done showing that a parallelogram with perpendicular diagonals is a rhombus. The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side. So let's review the properties of parallelograms - one of them is that the diagonals bisect each other. We now need either another side, or another angle to show the triangles are congruent. Prove that (i) ar (ABCD) ar (EFCD) Given that ABCD is a. We have one common side - OD, and since the diagonals are perpendicular, m∠AOD = m∠COD=90°, and so ∠AOD ≅ ∠COD. Example 1 In given figure, ABCD is a parallelogram and EFCD is a rectangle. We will do this using congruent triangles. So we need to show that one pair of adjacent sides is equal. We know that the opposite sides in ABCD are equal, because it is a parallelogram. StrategyĪBCD is a rhombus if all its sides are equal. ProblemĪBCD is a parallelogram, and its diagonals are perpendicular - AC⊥DB. Final answer: To prove that ABCD is a parallelogram, we need to show that opposite sides are equal in length and parallel to each other. Here we will show the converse- that if a parallelogram has perpendicular diagonals, it is a rhombus - all its sides are equal. We've seen that one of the properties of a rhombus is that its diagonals are perpendicular to each other. A rhombus is a special kind of parallelogram, in which all the sides are equal.
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