State the theorem which is used …

Construction-From center O draw a straight line connecting point of contact A on tangent to a point M on chorday

Proof-since angle between radius  and tangent at point of contact is 90 thus angle formed between  radius and tangent at A is 90

Now chord//tangent

Thus angle OMD is also 90

This proves that OM   is perpendicular to chord

A perpendicular from centre to chord bisects the chord

Thus CM=DMITRY

IN TRIANGLE ACM AND ADM 

CM=DMITRY

angleAMC=AND

AM=AM

THUS TRIANGLE ACM AND ADM ARE CONGURENT  thus AC=AD

AS TWO SIDES ARE EQUAL TRIANGLE ACD IS ISOSCELES 

Ans. 1) Section Formula : The Co-ordinates of the point P(x,y) which divides the line joining the point $A(x_1,y_1) and B(x_2,y_2)$ Internally in the Ration m:n  is given by $\left(\left(\frac{m{x}_2 +n{x}_1}{m+n}\right),\left(\frac{m{y}_2+n{y}_1}{m+n}\right)\right)$. 

2. Given : Chord CD is parallel to Tangent FAE.

To Prove : Triangle ACD is Isosceles.

Proof : $\angle CAF = \angle ACD . . . \left(1\right) (Alternate angles)$

$\angle CAF = \angle ADC . . . \left(2\right) (Alternate Segment theorem)$

$From \left(1\right) and \left(2\right) , we get$

$\angle ACD = \angle ADC \Rightarrow ∆ACD is isosceles triangle .$