Let \( ABC\) be a triangle with circumcentre \(O\). The points \(P\) and \( Q \)are interior points of the sides \(CA\) and \(AB\) respectively. Let \(K,L\) and \( M \)be the mid points of the segments \(BP,CQ\) and \(PQ\) respectively, and let \(\tau\) be the circle passing through \(K,L \)and \( M\).Suppose that the line \(PQ\) is tangent to the circle \(\tau\).Prove that \(OP=OQ\).

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TopNewestQP 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 – Anubhav Singh · 3 years, 10 months ago

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this is my favourite question – Anubhav Singh · 3 years, 7 months ago

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Hmm ... ..ho gaya. – Starwar Clone · 3 years, 10 months ago

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– Ankush Tiwari · 3 years, 10 months ago

HOW???Log in to reply

– Anubhav Singh · 3 years, 2 months ago

Dono bhai 95% pakar bhag gaye....mil jao kisi din bahut marenge.... :)Log in to reply

– Anubhav Singh · 3 years, 10 months ago

add a new status :)Log in to reply