Let ABCD be a tangential quadrilateral with inscribed circle k(O,r) which is tangent to the sides BC and AD at K and L, respectively. Show that the circle with diameter OC passes through the intersection point of KL and OD. Proposed by Ilija Jovchevski
Problem
Source: 2021 Junior Macedonian Mathematical Olympiad P2
Tags: geometry
08.06.2021 21:43
Lema: In △ABC where A′,B′,C′ are the foot of the altitudes from A,B,C respectively and H is the orthocenter. If we draw a circle ω with center H and radius HA′ that intersects the line B′C′ in D and E. Then D and E are the second tangency points from B and C to ω. Proof: Let (BC′H) intersect for the second time B′C′ in E′. Then \angle BC’H = \angle BE’H = 90º and \angle E’C’H = \angle E’BH. Because BC’B’C is cyclic \angle B’C’C = \angle B’BC which is equivalent to \angle E’C’H = \angle HBA’. Then \triangle BE’H and \triangle BA’H are congruent because they are similar by angles and share side BH, so HA’ = HE’ and E = E’, but as \angle BE’H = 90º then BE is tangent to \omega. Now if BE and DC intersects at K, H is the incenter of \triangle BKC. As the line DE (between two tangency points) intersects CH at C’ and \angle BC’H = 90º it belongs to the circle (BEH), which is a circle with diameter BH.
10.06.2021 15:21
Let AD and BC meet at E. Then k is incircle of \triangle EDC and L and K are touching point. From In-touch\ chord Lemma feet of perpendicular from C to DO is on LK and Let's say it R. Since \angle ORC=90 R\in circle with diameter OC
10.06.2021 15:40
Iran Lemma.