Question

OR
Prove that the sum of any two sides of a triangle is greater than the third side.​

Answer 1

Step-by-step explanation:

Construction: In \Delta ABCΔABC, extend ABAB to D in such a way that AD = ACAD=AC.

In \Delta DBCΔDBC, as the angles opposite to equal sides are always equal, so,

\angle ADC = \angle ACD∠ADC=∠ACD

Therefore,

\angle BCD > \angle BDC∠BCD>∠BDC

As the sides opposite to the greater angle is longer, so,

BD > BCBD>BC

AB + AD > BCAB+AD>BC

Since AD = ACAD=AC, then,

AB + AC > BCAB+AC>BC

Hence, sum of two sides of a triangle is always greater than the third side.

Answer 2

Answer:

If There Are Any Mistakes Please Make Sure You Correct It.

Step By Step Explanation:-

GIVEN

A Triangle ABC⇒ From Fig: 1

TO PROVE

AB+AC > BC.

CONSTRUCTION

Extent AB To D Such That AD = AC

PROOF⇒ Look Both Fig: 1 And 2

In Δ ACD

AD = AC ⇒ In Triangle Angles Opposite To Equal Sides Are Equal

So, ∠1 = ∠2

Now,

In Δ BCD > ∠2

BD > BC⇒ Sides Opposite To The Larger Angle Is Equal

AB + AD > BC⇒ BD = AB + AD

AB + AC > BC⇒ AD = AC By Construction

Hence Proved.