Show that of all lines segment …
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Sia ? 5 years, 8 months ago
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∘ (Angle sum property of triangle)
∠P+90∘+∠M=180∘ (As PN ⊥MN, ∠N=90∘
∠P+∠M=180∘−90∘
∠P+∠M=90∘
Since angle can't be 0 or negative
Hence,
∠M<90∘
∠M<∠N
PN < PM (Side opposite to the greater angle is longer)
∴ Perpendicular line segment is the shortest.
Hence proved.
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