Let be a triangle with circumcentre . The points and are interior points of the sides and respectively. Let and be the mid points of the segments and respectively, and let be the circle passing through and .Suppose that the line is tangent to the circle .Prove that .
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QP tangents with the small circle at M, we have ∠QMK = ∠MLK.
M, K and L are midpoints of PQ, BP and QC, respectively; therefore,
KM || QB, KM = ½ QB(a), ML || PC, ML = ½ PC(b)
and ∠QMK = ∠MQA, or ∠MLK = ∠MQA.
ML || PC and KM || QB therefore ∠QAP = ∠KML
The two triangles QAP and KML are similar since their respective angles are equal. Therefore, ML/QA = KM/AP
From (a) and (b) AP × PC = QA × QB(c)
Extend PQ and QP to meet the larger circle at U and V, respectively.
In the larger circle UV intercepts AB at Q, we have
QU × QV = QA × QB or QU × (QP + PV) = QA × QB (i)
UV intercepts AC at P, we have UP × PV = AP × PC or (QU + QP) × PV = AP x PC (ii)
From (i) and from (c) QU × (QP + PV) = AP × PC
Therefore, from (ii) QU × (QP + PV) = (QU + QP) × PV
Or PV = QU and M is also the midpoint of UV and OM ⊥UV
Therefore OP = OQ
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