Question

In the given figure, sides AB and AC of AABC are extended to points P and Q, respectively. Also, ZPBC < ZQCB., Show that AC > AB.​

Answer 1

Answer:

Given: ∆ABC is a Triangle &

∠PBC < ∠QCB

To Prove:

AC>AB

Proof:

∠ABC + ∠PBC = 180° ( by linear pair axiom)

⇒ ∠ABC = 180° – ∠PBC

also,

∠ACB + ∠QCB = 180° (by Linear Pair axiom)

⇒ ∠ACB = 180° – ∠QCB

Since,

∠PBC < ∠QCB therefore, ∠ABC > ∠ACB

Here, the side opposite to the larger angle is longer.

Hence, AC > AB

Step-by-step explanation: