Please do part(c) proving.Thanks alot.
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.