Question from http://www.sgforums.com/forums/2297/topics/368877
In the diagram, P is any point on the semicircle centre O, and PQ is perpendicular to AB. The inscribed circle centre C touches PQ, AB and the semicircle at D, E and F respectively. Prove that
(a) A,D and F lie on the same straight line
(b) AD * AF = AQ * AB
(c) AE2 = AQ2 + AQ * QB
(d) AE = AP
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Answer:
(a) DC is perpendicular to PQ, hence DC is parallel to AO.
OCF is a straight line.
=> angle DCF = angle AOF
=> FDC and FAO are similar triangles.
Hence, A, D and F lies on the same straight line.
(b) Draw a straight line from F to B.
angle AFB = 90 degrees = angle AQD
angle DAQ = angle BAF
=> angle ADQ = angle ABF
=> ADQ and ABF are similar triangles.
=> AD / AB = AQ / AF
AD X AF = AQ X AB (proved)
(c)
AE2 = AF * AD( tangent-secant theorem)
From part (b) AD X AF = AQ X AB
AE2 = AQ * AB
AE2 = AQ ( AQ + AB)
AE2 = AQ2 + AQ * AB (proven)
(d) Draw a full circle and extend a line to form a chord
AD = AQ * QD
PQ = QZ ( radius from centre bisect chord)
PQ * QZ = PQ2
Therefore, PQ2 = AQ * QB ( Intersecting chords theorem)
From part (c) : AQ2 = AE2 - (AQ)(QB)
AP2 = AQ2 + PQ2
Therefore, AP2 = [AE2 - (AQ)(QB)] + [(AQ)(QB)] ( Replaced found eqn)
AP2 = AE2
AP = AE (proved)
Thanks to forumers at sgforums for providing the question and answer :D
