Question

12. P is any point inside the triangle ABC. Prove
that: ZBPC > ZBAC.​

Answer 1

Answer:

see below

Step-by-step explanation:

as we know....

PBC > ABC & PCB > ACB hence

PBC + PCB > ACB + ABC....(1)

now PBC + PCB +BPC = ABC + BCA + BAC = 180°

(PBC + PCB) + BPC = (ACB + ABC) + BAC

now using (1),

ZBPC > ZBAC.

Note :

3+7 =4+6 then 3<4 so 7>6