Question

let ABC be a triangle with circumcentre O. the points P and Q are interior points of the sides CA and AB. let K, L, M be midpoint of BP, CQ, PQ. and let T be the circle passing through K, L, M. supposed that line PQ is tangent to circle T. prove OP=OQ

Answer 1

Whne we look at QP tangents with the small circle at M

We will have ∠QMK = ∠MLK.

M, K and L are midpoints of PQ, BP and QC

KM || QB, KM = ½ QB (*), ML || PC, ML = ½ PC (**)

and ∠QMK = ∠MQA, or ∠MLK = ∠MQA.

ML || PC and KM || QB therefore ∠QAP = ∠KML

QAP and KML are triangles which are similar because their angles are equal.

ML/QA = KM/AP

From (*) and (**) AP × PC = QA × QB (***)

We will extend PQ and QP to meet the larger circle at U and V

In the larger circle UV intercepts AB at Q, we will have

QU × QV = QA × QB or QU × (QP + PV) = QA × QB (i)

UV intercepts AC at P, we will have UP × PV = AP × PC or (QU + QP) × PV = AP x PC (ii)

From (i) and (***) QU × (QP + PV) = AP × PC

From (ii) QU × (QP + PV) = (QU + QP) × PV

Or PV = QU and M is also the midpoint of UV and OM ⊥UV

Hence, OP = OQ