Let \(ABC\) be a right-angled triangle with \(\angle B= 90^\circ\). Let \(BD\) be the altitude from \(B\) on to \(AC\). Let \(P, Q\) and \(I\) be the incenters of triangles \(ABD, CBD\) and \(ABC\), respectively. Show that the circumcentre of the triangle \(PIQ\) lies on the hypothenuse \(AC\).

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TopNewestWe begin with the following lemma: Lemma: Let XY Z be a triangle with \XY Z = 90 + . Construct an isosceles triangle XEZ, externally on the side XZ, with base angle . Then E is the circumcentre of 4XY Z. Proof of the Lemma: Draw ED ? XZ. Then DE is the perpendicular bisector of XZ. We also observe that \XED = \ZED = 90 . Observe that E is on the perpendicular bisector of XZ. Construct the circumcircle of XY Z. Draw perpendicular bisector of XY and let it meet DE in F. Then F is the circumcentre of 4XY Z. Join XF. Then \XFD = 90. But we know that \XED = 90. Hence E = F. Let r1, r2 and r be the inradii of the triangles ABD, CBD and ABC respectively. Join PD and DQ. Observe that \PDQ = 90. Hence PQ2 = PD2 + DQ2 = 2r2 1 + 2r2 2: Let s1 = (AB + BD + DA)=2. Observe that BD = ca=b and AD = p AB2 BD2 = q c2 ca b 2 = c2=b. This gives s1 = cs=b. But r1 = s1 c = (c=b)(s b) = cr=b. Similarly, r2 = ar=b. Hence PQ2 = 2r2 c2 + a2 b2 = 2r2: Consider 4PIQ. Observe that \PIQ = 90+(B=2) = 135. Hence PQ subtends 90 on the circumference of the circumcircle of 4PIQ. But we have seen that \PDQ = 90. Now construct a circle with PQ as diameter. Let it cut AC again in K. It follows that \PKQ = 90 and the points P;D;K;Q are concyclic. We also notice \KPQ = \KDQ = 45 and \PQK = \PDA = 45. Thus PKQ is an isosceles right-angled triangle with KP = KQ. Therfore KP2 + KQ2 = PQ2 = 2r2 and hence KP = KQ = r. Now \PIQ = 90 + 45 and \PKQ = 2 45 = 90 with KP = KQ = r. Hence K is the circumcentre of 4PIQ. (Incidentally, This also shows that KI = r and hence K is the point of contact of the incircle of 4ABC with AC.) – Saikat Sengupta · 1 month, 4 weeks ago

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