Question

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

Answer 1

Answer:

Proof is below.

Step-by-step 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.