Question

Show that (AB+BC)+CD = AB+(BC+CD) by using parallelogram method.
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Answer 1

Step-by-step explanation:

In quadrilateral ABCD , draw the diagonals AC and BD which meet at point P.

we know that sum of any two sides of a triangle is always greater than the third

side. In ∆PAB. , PA+PB > AB………………..(1)

In ∆ PBC. , PB+PC > BC………………………..(2).

In. ∆ PCA. , PC+PA > CA……………………….(3).

In ∆ PAD. , PA +PD > DA………………………(4).

Adding (1) , (2) , (3) and (4).

(PA+PB+PC+PD) > AB+BC+CD+DA..

{(PA+PC)+(PB+PD)} > AB+BC+CD+DA.

{ AC+BD} > AB+BC+CD+DA. Proved.

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