Referring to the attached pic:

APQ is a semi-circle with PQ as the diameter.

QP is extended to B, and BA is a tangent to the circle at A.

AC is perpendicular to PQ and PC:CQ = 1:4.

AB = 8cm

(a) Prove $\displaystyle AC^2 = (PC)(QC)$

(b) Prove $\displaystyle AB^2 = (BP)(BQ)$

(c) If BP = y and PC = k, prove that $\displaystyle k = \frac{3}{5} y$

(d) Find BP

(e) Find the radius of circle.

My work:

(a)

triangles APC ~ ACQ (same angles)

hence AC:PC = QC:AC

$\displaystyle AC^2 = (PC)(QC)$

(b)

triangles BAQ ~ BPA (same angles)

hence AB:BQ = BP:AB

$\displaystyle AB^2 = (BP)(BQ)$

(c)

since PC:CQ = 1:4,

CQ = 4k and hence PQ = 5k

hence $\displaystyle \frac{k}{AC} = \frac{AC}{4k}$

$\displaystyle AC^2 = 4k^2 = (2k)^2$

hence we have AC = 2k

$\displaystyle AP^2 = AC^2 + PC^2$ (Pythagarus)

$\displaystyle AP^2 = (2k)^2 + k^2$

$\displaystyle AP^2 = 5k^2$

AP = (k)(5^0.5)

Also $\displaystyle AQ^2 = PQ^2 - AP^2$

$\displaystyle AQ^2 = (5k)^2 - 5k^2$

$\displaystyle AQ^2 = 20k^2$

AQ = (2k)(5^0.5)

y:8 = AP:AQ = k(5^0.5):2k(5^0.5)

y:8 = 1:2

hence y = 4

and 8:BQ = AP:AQ = 1:2

8:(5k + y) = 1:2

8:(5k + 4) = 1:2

16 = 5k + 4

k = $\displaystyle \frac{12}{5}$

Therefore, $\displaystyle \frac{3}{5} y = (\frac{3}{5})(4) = \frac{12}{5} = k$

Confused here: relation between k and y is supposed to be proved without working out their values in advance. Is there another method to prove that k = 3y/5 ????

(d)

BP = y = 4cm

(e)

PQ = 5k = 12cm