M-is-the-midpoint-of-segment-AB-Prove-that-for-every-point-P-in-space-PM-PA-PB-2-

Question Number 64203 by Prithwish sen last updated on 15/Jul/19

M is the midpoint of segment AB . Prove

that, for every point P in space,

∣ PM ∣ ⩽ \frac{∣ PA ∣ + ∣ PB ∣}{2}

Commented by Prithwish sen last updated on 16/Jul/19

Sir, {isn}^{'} t PB or PA will be the diagonl if

triangle PAB transform into a parallalogram .

Commented by MJS last updated on 16/Jul/19

draw A B and M

now choose a point P and its mirror P^{'} so that

M is also the mid of {P P}^{'}

\Rightarrow

A B is one diagonal and {P P}^{'} is the other one

∣ {P P}^{'} ∣= 2 ∣ P M ∣⩽∣ P A ∣ + ∣ P B ∣

Commented by Prithwish sen last updated on 16/Jul/19

∵ {PAP}^{'} is a triangle . Ok sir thank you .

Commented by MJS last updated on 16/Jul/19

{you}^{'} re welcome

sorry I have no possibility to add drawings

at the moment \dots

Commented by Prithwish sen last updated on 16/Jul/19

No sir you have already done a lot . I appreciate

your effort Thank you very much again .

Answered by MJS last updated on 16/Jul/19

think of a triangle

sides A B, P A, P B

now make it a parallelogram

obviously 2 ∣ P M ∣ which is one diagonal

must be shorter than the sum of two sides

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