Question involving area and perimeter of two parallelograms sharing a diagonal.

Question

Given two parallelograms $P1$ and $P2$ sharing a diagonal, such that area of $P1$ is greater than area of $P2$, can we say that the perimeter of $P1$ is greater than the perimeter of $P2$ ?

Actually I was trying to prove another question : Proof of a geometric statement

I thought of using this argument as shown in the following picture . Here $1$ represents parallelogram with sides $AB$ and $AC$ and $2$ represents area of parallelogram with sides $BD$ and $CD$. :

Answer 1

Since I can not comment - Your question boils down to a simpler claim: The perimeter of a triangle created by connecting a point in a scalene and acute triangle to any two vertices, is smaller than the original triangle perimeter.

I tested this with Geogebra - and it seems correct. The comment made by Sawarnik is not relevant to your claim - which is true but require proof.

I provided a proof on your first post, that the parallelogram that contains the smaller one has a larger perimeter.

You may accept the answers on both posts. Thanks.

Answer 2

Try the following two parallelograms:

(-2,0),(0,1.01),(2,0),(0,-1.01)

(-2,0),(-2,1),(2,0),(2,-1)