#### Question

In the given figure sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

#### Solution

In the given figure,

∠ABC + ∠PBC = 180° (Linear pair)

⇒ ∠ABC = 180° − ∠PBC ... (1)

Also,

∠ACB + ∠QCB = 180°

∠ACB = 180° − ∠QCB … (2)

As ∠PBC < ∠QCB,

⇒ 180º − ∠PBC > 180º − ∠QCB

⇒ ∠ABC > ∠ACB [From equations (1) and (2)]

⇒ AC > AB (Side opposite to the larger angle is larger.)

Is there an error in this question or solution?

Solution In the given figure sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB. Concept: Inequalities in a Triangle.