show angle APB = 60 degree, if AD & BC are produced to meet at P

Question

in a circle, with centre O, a cyclic quadrilateral ABCD is drawn with AB as a diameter of the circle and CD equal to radius of the circle.

Answer 1

By convention, a quadrilateral named ABCD has vertices named A, B, C, and D, in that order. So, if AB is the base and DC is the top, D is "above" A and C is "above" B. I'll assume that's the case (though the problem still works if C and D are switched).

 

 

Go ahead and draw a picture of the situation. You'll have a diameter of the circle called AB which passes through its center O. Make this a horizontal line. Then, you'll have a segment half that length (ie., the length of the radius) called DC. Draw this as a roughly horizontal line above AB, where D is above A and C is above B.

 

Now, extend the lines AD and BC until they meet. Call that point P. It should be above and outside of the circle. Finally, draw the radii OD and OC.

 

 

First, notice that arc AB (the arc that doesn't include C and D) is exactly half the circle, or 180°.

 

Also, DOC is an equilateral triangle: all three of its sides are the length of the circle's radius. So, <DOC is 60°. Since <DOC is a central angle, its corresponding minor arc DC is also 60°.

 

Now, we'll use the formula relating an angle with vertex outside the circle (<APB) and its two intercepted arcs (DC and AB):

 

angle = (1/2)(larger arc - smaller arc)

m<APB = (1/2)(AB - DC)

m<APB = (1/2)(180 - 60)

m<APB = (1/2)(120)

m<APB = 60

 

So, <APB = 60°, like we wanted.