Ex 7.4, 6 - Show that of all line segments drawn from a - Side inequality

Ex 7.4, 6 - Chapter 7 Class 9 Triangles - Part 2

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Ex7.4, 6 Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest. Let the given point be P and line be l We draw two line segments PN & PM such that PM ⊥ MN We have to prove: PM > PN In ΔPNM, ∠P + ∠N + ∠M = 180° ∠P + 90° + ∠M = 180° ∠P + ∠M = 180° – 90° ∠P + ∠M = 90° Since angle can’t be 0 or negative Hence, ∠ M < 90° ∠ M < ∠ N PN < PM ∴ Perpendicular line segment is the shortest Hence proved

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