Theorem 8.6 - Class 9 - Diagonals of a parallelogram bisect each other
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Theorem 8.6 - Class 9 - Diagonals of a parallelogram bisect each other
Theorem 8.6 The diagonals of a parallelogram bisect each other Given : ABCD is a Parallelogram with AC and BD diagonals & O is the point of intersection of AC and BD To Prove : OA = OC & OB = OD Proof : Since, opposite sides of Parallelogram are parallel.
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Does the diagonal actually exist? 📐🤯 Imagine a perfectly square park, 100ft by 100ft. You need to get to the opposite corner. 🔹 Euclidean Math: Walk the diagonal = 141 ft. 🔹 Grid Rules: You can only move North or East. If you take the edge, it’s 200 ft. Too long. So you cut through the middle using stairs. 🪜 North, East, North, East. It looks shorter. So you shrink the stairs. You make them microscopic. You break the
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Does the diagonal actually exist? 📐🤯 Imagine a perfectly square park, 100ft by 100ft. You need to get to the opposite corner. 🔹 Euclidean Math: Walk the diagonal = 141 ft. 🔹 Grid Rules: You can only move North or East. If you take the edge, it’s 200 ft. Too long. So you cut through the middle using stairs. 🪜 North, East, North, East. It looks shorter. So you shrink the stairs. You make them microscopic. You break the
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