Question

In the figure, sides AB and AC of $\Delta {\rm{ABC}}$ are extended to point P and Q respectively. Also, \[\angle {\rm{PBC}} < \angle {\rm{QCB}}\]. Show that$AC > AB$.

Answer 1

Hint: When two lines intersect then linear pair if formed. The concept of linear angle property is that the addition of adjacent angles is equal to $180^\circ $ on a straight line. Use liner angle property to solve this question. Also, use the theorem that the opposite side of a larger angle is larger and the opposite side of the smallest angle is small.

Complete Step-by-step Solution
Complete step-by-step solution:
Given, In $\Delta {\rm{ABC}}$ sides AB and AC are extended up to point P and Q respectively and\[\angle {\rm{PBC}} < \angle {\rm{QCB}}\].
We can assume the angles as $\angle ABC = \angle 1$,$\angle PBC = \angle 2$, $\angle ACB = \angle 3$and $\angle QCB = \angle 4$.
Given values are, \[\angle {\rm{PBC}} < \angle {\rm{QCB}}\] which means that \[\angle {\rm{2}} < \angle {\rm{4}}\].
We know the linear property of angles says that the sum of adjacent angles form on a straight line is $180^\circ $. Use this rule for angles \[\angle 1\], \[\angle 2\], \[\angle 3\,\] and \[\angle 4\].
$\begin{array}{c}
\angle 1 + \angle 2 = 180^\circ \\
\angle 2 = 180^\circ - \angle 1
\end{array}$…….(1)
$\begin{array}{c}
\angle 3 + \angle 4 = 180^\circ \\
\angle 4 = 180^\circ - \angle 3
\end{array}$…….(2)

Now, the relation between angle 2 and angle 4 is,
\[\angle {\rm{2}} < \angle {\rm{4}}\]
On simplifying the above equation (1) and equation (2) we get the values as:
$\begin{array}{c}
180^\circ - \angle 1 < 180^\circ - \angle 3\\
 - \angle 1 < - \angle 3
\end{array}$
Here, the negative of $\angle 1$ is less than the negative of $\angle 3$, if we remove this negative sign then positive $\angle 1$ becomes greater than $\angle 3$.
$\angle 1 > \angle 3$
The side opposite to the$\angle 1$ should be greater than the side opposite to angle $\angle 3$.
Therefore, ${\rm{AC > AB}}$

Note: In such types of solutions there is a tricky part to solve it that is when removing negative signs from less than angles it makes the opposite effect and makes greater than sign between the angles.