But question 25 is very interesting. Try to solve it.

( 1)

ADC=90, then, because CAD=45 and ABC=90,

Because BAC=30 and AC=2, then CD= root 2.

(2)

Because ADC = ABC = 90, ABCD is on the same circle and AC is diameter =2.

If BD's circumferential angle BAD = 75, then the central angle B0D= 150, BD = ACCOS15 = 2cos15.

cos 15=sicos(30/2)

= root (( 1+cos30)/2)

= root (1+cos30)/ root 2

BD=2cos 15= root 2* root (1+cos30)

BD= root 2* root (BC+AB)

(3)

If ADC=60 and CAD=ACD=60.

Then ABC= 120, BAC=30, then BCA = 30-, AB=BC, BAD=ABD=90.

So ABCD is a circle on the diameter BD, ABD=60.

So BD=2AB=2BC=2.

The second question is the most interesting.