5. In circle K shown, secant PEC intersects chord BKD at F. If AEAB, MED=32", and
mBC:mCD = 2:1, then find the following:
B
4
1120
(a) mZP
1+2=3
3 x=180
3 3
x=60
(b) mZCFD
60X2=120 BC=120
180-32=148
| MLP=23
148/2=74
Ź(120-74)=23
к
174
F
E
32
D

it 2

step-by-step explanation:

8^1/3=2

the answer is 11.2 (i know i’m late)

step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)