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# Question on circles

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 $$OP=OQ$$.

Note by Aman Tiwari
3 years, 10 months ago

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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 · 3 years, 10 months ago

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

Hmm ... ..ho gaya. · 3 years, 10 months ago

HOW??? · 3 years, 10 months ago

Dono bhai 95% pakar bhag gaye....mil jao kisi din bahut marenge.... :) · 3 years, 2 months ago