Prove The Following
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Below.

Step-by-step explanation:

m < PCA = m  < CAD ( alternate angles)

m < BAD = m <CAD ( given)

Also m < CAB = m < PCA + m < CPA  ( exterior angle of a triangle theorem).

Also m < CAB = m< CAD + m < BAD (given)

So  m < PCA + m < CPA =  m < CAD + m < BAD

But m < PCA = m  < CAD

so  m < CAD + m < CPA = m < CAD + m BAD

So m < CPA = m < BAD

But m < PCA = m < CAD = m BAD

therefore m < CPA = m < PCA

That makes triangle CAP isosceles

Therefore AP = AC.

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b