Showing The Sum Of Distances From Parallelogram Vertices To Any Interior Point Is Less Than Perimeter

Question

Let ABCD is a parallelogram and P be any interior point in ABCD.

Here is my attempt to show |PA|+|PB|+|PC|+|PD|< 2(|AB|+|BC|) so far:

$ |PA|+|PB|<|AD|+|BD| \\ |PA|+|PD|<|AB|+|BD| \\ |PD|+|PB|<|AB|+|AD|\\ |PA|+|PC|<|AD|+|DC|\\ |PA|+|PD|<|AB|+|BD|\\ |PD|+|PC|<|AD|+|AC|\\ |PB|+|PC|<|AC|+|BD|\\ |PB|+|PC|<|AC|+|BD| $

adding all together

$ \implies 4(|PA|+|PB|+|PC|+|PD|)< 4(|AB|+|AD|+|BD|+|AC|)$

$ \implies |PA|+|PB|+|PC|+|PD|< |AB|+|AD|+|BC|+|AC|$

$ \implies |PA|+|PB|+|PC|+|PD|< |AB|+|AD|+|AB|+|AD|+|AB|+|AD|$

$ \implies |PA|+|PB|+|PC|+|PD|< 3(|AB|+|AD|)$

and couldn't show that less than $2(|AB|+|AD|)$.

i also showed that

$2(|PA|+|PB|+|PC|+|PD|)>2(|AB|+|AC|)>|AC|+|BD|<|AB|+|BC|$

but it doesn't help. Thanks in advance for guidences and tips :)

Answer 1

Construct the above diagram. Then, apply the triangle inequalities

$$DP < PD' + D'D, \>\>\>\>\>DP' < DA' + A'P' \>\>\>\>\>CP' < CB' + B'P' \>\>\>\>\>CP < CC' + C'P$$

Add them up to get

$$DP + DP' + CP' + CP $$ $$< (PD' + C'P) + (D'D + DA') + ( A'P' + B'P') +( CB'+ CC') $$

Thus,

$$DP + AP + CP + BP < 2(AB + BC) $$

Answer 2

Use the following for the two diagonal paths:

If $P$ is an interior point of triangle $ABC$, then $|AP|+|PB|<|AC|+|CB|$.

To see this, thge following image shows the "worst case" where $|AP|=|AQ|<|AC|+|CQ|$ and $|PB|=|RB|$.