Find this important question of ncert class 9th of chapter triangles .Sir please help me to find out the easiest and simplest solution of question 2 of exercise 7.4, give me the best solution of this question. In Fig. 7.48, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, PBC < QCB. Show that AC > AB.
It is given that PBC < QCB
We know that ABC + PBC = 180°
So, ABC = 180°-PBC
Also,
ACB +QCB = 180°
Therefore ACB = 180° -QCB
Now, since PBC < QCB,
∴ ABC > ACB
Hence, AC > AB as sides opposite to the larger angle is always larger