wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Show that difference of any two sides of a triangle is less than to the third side.

Open in App
Solution



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,

So, p = p

Now, let

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.
(equation 1)
Now, using exterior angle property in triangle BCD,
(equation 2)
Now, by comparing equation 1 and 2,
So, y > x

BC > CD

CD < BC

AC - AD < BC

Since, AD = AB

Therefore, AC - AB < BC

Hence, proved.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Triangle Properties
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon