Question

${\underline{\sf \green{Question:-}} }$
Explain the proof of :-
In any triangle the sum of any two sides is greater than the third side.​

Answer 1

Given That: -

• ↠ A triangle ( ️) PQR

What we have to prove: -

• ↠ The sum of any sides is greater then the third side. or ∠PQ + ∠PR ≻ ∠QR

Construction:-

1. Extend PR to a point S such that RS = RQ
2. join RS

Let's proof :-

In triangle PQR , we have

• ⇝ QR = SR (By construction)

By theorem : In triangle angles opposite to equal side are equal

Therefore ∠1 = ∠2 ( ∠PRS =∠PSR )

Now ,

∠QRS =∠QRP +∠1

• QRP +∠2 ( ∠1 =∠2)

∠QRP > ∠2

Now in triangle QRS

∠QRS > QR

⇝QS > QR ( In triangle side opposite to a larger angle is longer)

⇝PQ + PS > QR (QS = PQ + PS )

⇝PQ + PR > QR (PS = PR By construction)

Similarly we can prove

⇝ PQ + QR > PS & PR + QR > PQ