The sum of lengths of any two sides of a triangle is always ………………………….the third side.
A.Greater than
B.Less than
C.Equal to
D.None of these

Hint: Assume a \[\Delta ABC\] and then extend the side BA to D such that \[AD=AC\] . If two sides are equal then the angles opposite to them are also equal. So, \[\angle ADC=\angle ACD\] . In the \[\Delta BCD\] , we have \[\angle BCD>\angle ACD\], but \[\angle ADC=\angle ACD\]. So, \[\angle BCD>\angle ADC\] . We know the theorem that the side opposite to the larger angle is longer. So, \[BD>BC\] . BD is the summation of the side AB and AD. But we have \[AD=AC\] . Now, solve it further.

Complete step-by-step answer:


First of all, let us assume a \[\Delta ABC\]and then extend the side BA to D such that \[AD=AC\] . Now, join D to C.
In the \[\Delta ADC\] , we have \[AD=AC\] ……………….(1)
We know the theorem that the angles opposite to equal sides are also equal.
We have \[\angle ADC\] and \[\angle ACD\]opposite to the sides AD and AC respectively.
So, \[\angle ADC=\angle ACD\] ………………….(2)
In the \[\Delta BCD\] we have,
\[\angle BCD=\angle ACD+\angle ACB\] …………………..(3)
From equation (2) and equation (3), we get
\[\angle BCD=\angle ADC+\angle ACB\]
Thus, we can write \[\angle BCD>\angle ADC\] .
We know the theorem that the side opposite to the larger angle is longer.
We have the side BD and BC opposite to the angles \[\angle BCD\] and \[\angle ADC\] respectively.
So, \[BD>BC\] ……………..(4)
From the figure, we can see that the side BD is the summation of the side BA and AD.
Transforming equation (4), we get
 \[BD>BC\]
\[BA+AD>BC\] ……………………..(5)
From equation (1) and equation (5), we get
\[BA+AC>BC\] .
The summation of the two sides is greater than the third side.
Hence, option (A) is the correct option.

Note: In this question, one may think that why only the side BA is extended to D. The answer for this is we can extend any side to D but we have to take one side equal to the other side. Let us understand with an example.


Here, we are extending the side CB to D such that AB=BD. Now using the same approach, solve this.