Question

state and prove theorem of angle in alternate segment

Answer 1

The alternate segment theorem (also known as the tangent-chord theorem) states that in any circle, the angle between a chord and a tangentthrough one of the end points of the chord is equal to the angle in the alternate segment.

The chord DF divides the circle into two segments, and we're interested in the angle between this chord and the tangent at D, and the angle in the other (alternate) segment, E. 

We need to show that D=E (see fig 1)


As before, the first step is to draw in radiuses from points on the circumference to the centre, A. 




In the second diagram, fig 2, it is clear that the three triangles ADF, AFE and AED are isosceles, as a pair of sides in each triangle are radiuses. 
Thus the two angles in ADF marked 'x' are equal (and similarly for y and z in the other triangles.) 

Now, at D, α=90 - x 
(angle between radius and tangent is right angle) 

In triangle DEF, (x+y) + (y+z) + (z+x) = 180
2(x+y+z) = 180 
x+y+z = 90 
but at E, β = y+z = 90 - x =α 

Answer 2

Answer:

A chemical equation should be written with the reactants (if there are two or more should be separated by '+' symbol, same applicable to products) on the left side of an arrow and the products of the chemical reaction on the right side of the equation.

Step-by-step explanation: