Question

Given △ABC, AB=AC, P is interior of △ABC, PC>PB Prove: m∠PCA>m∠PBA

Answer 1

Proof is below.

Explanation:

If two sides are equal, then angles opposite to them are also equal.

The angle opposite to the greater side is greater than the angle opposite to lesser side.

Given:

In
`\Delta ABC`,

AB = AC

PC > PB

As sides AB = AC,

∴
`m\angle ABC=m\angle ACB=a(Let)`

As PC > PB, then, from the theorem of greater angle lies opposite to the greater side,

∴
`m\angle PBC > m\angle PCB`

Let
`m\angle PBC =x\textrm{ and } m\angle PCB=y`

So, angle PBA is,
`m\angle PBA=m\angle ABC-m\angle PBC=a-x`

Angle PCA is,
`m\angle PCA=m\angle ACB-m\angle PCB=a-y`

Now, we have,
`x > y`

Multiply by -1 both sides. This changes the inequality sign.

⇒
`-x < -y`

Adding
`a` on both sides, we get

`a-x<a-y`

But,
`m\angle PBA=a-x` and
`m\angle PCA=a-y`.

∴
`a-x<a-y\\m\angle PBA<m\angle PCA\textrm{ or}\\m\angle PCA>m\angle PBA`. Hence, it is proved.