Part c: Prove (BE)^2 = (AB)(BC)

In right triangle ABE, by Pythagorean theorem,

(BE)^2 = (AE)^2 -(AB)^2 ----------(1)

In right triangle CBE,

(BE)^2 = (EC)^2 -(BC)^2 ----------(2)

Add (1) and (2),

2(BE)^2 = [(AE)^2 +(EC)^2] -[(AB)^2 +(BC)^2] ---------(3)

Since AE is tangent to circle Q, then AE is perpendicular to EC, hence, triangle AEC is a right triangle.

So,

(AC)^2 = (AE)^2 +(EC)^2 -----------------(4)

Substitute that into (3),

2(BE)^2 = (AC)^2 -[(AB)^2 +(BC)^2] ------(5)

In the straight line ABC,

(AC)^2 = (AB +BC)^2

(AC)^2 = (AB)^2 +2(AB)(BC) +(BC)^2 --------(6)

Substitute that into (5),

2(BE)^2 = [(AB)^2 +2(AB)(BC) +(BC)^2] -[(AB)^2 +(BC)^2]

2(BE)^2 = 2(AB)(BC)

(BE)^2 = (AB)(BC)

Proven.