Question

Prove that in a triangle the difference of any two
sides is less than the third.
​

Answer 1

IT CAN BE PROVE IN ISOSCELES TRIANGLE.

Answer 2

Solution :

We have to prove that the difference of two sides of a triangle is less than the third side.

Refer to the attached image.

To prove:

• AC - AB < BC

Construction:

Mark a point 'D' on AC

such that the distance AB=AD

Therefore, we have to prove that AC - AD < BC

So, we have to prove that CD < BC

Proof:

Consider triangle ABD,

Since AB=AD

Since, opposite angles opposite to the equal opposite sides are always equal.

So, ∠ABD = ∠ADB

So, p = p

Now, let ∠DBC = x, ∠BDC = y

Now, using exterior angle property in triangle ABD,

Exterior angle property of a triangle states that the measure of an exterior angle is equal to the sum of the two interior angles.

y = p+ ∠A (equation 1)

Now, using exterior angle property in triangle BCD,

p = t +∠C

x =P - ∠C (equation 2)

Now, by comparing equation 1 and 2,

p + ∠A > p-∠C

So, y > x

BC > CD

CD< BC

AC- AD < BC

Since, AD = AB

Therefore, AC -AB < BC

Hence, proved.