Question

In a triangle, show that any side is greater than the difference of the other two sides. [Hint: Construct PS = PR ∴ ∠a = ∠b

In ΔQSR, ext. ∠a > ∠d ⇒ ∠b > ∠d

In ΔPSR, ext. ∠c > ∠b

∠c > ∠a ⇒ ∠c > ∠d

∴ QR > QS

⇒ QR > PQ – PS

QR > PQ – PR]

Answer 1

Here’s the step-by-step explanation to show that in a triangle, any side is greater than the difference of the other two sides, based on the provided hint:

Statement:

In a triangle, any side is greater than the difference of the other two sides.

We’ll prove:

For triangle PQR,

QR > |PQ − PR|

Construction (from hint):

Construct triangle PQR.

On side PQ, mark a point S such that PS = PR.

Now join SR and QR.

So in triangle PSR, PS = PR ⇒ It’s isosceles, so:

∠a = ∠b

Step 1:

In triangle QSR,

Exterior angle ∠a > Interior opposite angle ∠d 

But since ∠a = ∠b,

⇒ ∠b > ∠d

Step 2:

In triangle PSR,

Exterior angle ∠c > Opposite interior angle ∠b

⇒ ∠c > ∠b 

So, from step 1:

∠c > ∠b > ∠d 

⇒ ∠c > ∠d

Step 3:

In triangle PQR, comparing ∠c and ∠d:

Since ∠c > ∠d ⇒ side opposite ∠c (which is QR) is greater than side opposite ∠d (which is QS):

⇒ QR > QS

Step 4:

From the construction:

QS = PQ – PS and since PS = PR, 

⇒ QS = PQ – PR 

Thus, QR > QS = PQ – PR

Final Result:

QR > PQ – PR

So, any side (QR) is greater than the difference of the other two sides (PQ – PR).

Similarly, you can prove:

• PQ > |QR – PR| 
• PR > [PQ – QR|

This proves the triangle inequality in difference form.