×

Geometry

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 $$AC$$.

Note by Saikat Sengupta
7 months, 4 weeks ago

Sort by:

We 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.) · 7 months, 4 weeks ago

This one is the official solution. Use some constructions and a $$short$$ three stepped solution will come to you on its own feet... No need of such lemmas.If you need one I can write mine shorter solution.... . · 5 months, 2 weeks ago