Question on circles

Let ABC ABC be a triangle with circumcentre OO. The points PP and Q Q are interior points of the sides CACA and ABAB respectively. Let K,LK,L and M M be the mid points of the segments BP,CQBP,CQ and PQPQ respectively, and let τ\tau be the circle passing through K,LK,L and M M.Suppose that the line PQPQ is tangent to the circle τ\tau.Prove that OP=OQOP=OQ.

#Geometry #Proofs #MathProblem #Math

Note by Aman Tiwari
7 years, 8 months ago

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6 votes

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Comments

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

Anubhav Singh - 7 years, 7 months ago

this is my favourite question

Anubhav Singh - 7 years, 5 months ago

Hmm ... ..ho gaya.

Ujjwal Mani Tripathi - 7 years, 8 months ago

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HOW???

Ankush Tiwari - 7 years, 8 months ago

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

Anubhav Singh - 6 years, 11 months ago

add a new status :)

Anubhav Singh - 7 years, 7 months ago
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