In triangle ABC,∠BAC=120o. Let the angle bisectors of angles A;Band C meet the opposite sides at D;E andF respectively. Prove that the circle on diameter EF passes through D.
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Tags: geometry, circumcircle, geometry proposed
28.05.2008 14:54
We should only prove that ∠EDF=90∘. Lemma. Let ABC be a triangle with ∠A=60∘. BD is bisector of ∠B. A line, which passes through D intersect side AB at F. Then ∠FDB=30∘⇒CF is bisector of ∠C.
It is clear that ∠FIB=30∘. Apply Lemma in triangle ADB for bisector BI, line FI and obtain that DF is bisector of angle BDA. By analogy DE is bisector of angle ADC. So ∠FDE=12(∠BDA+∠CDA)=90∘.
29.05.2008 17:01
see here
30.05.2008 01:49
IWW wrote: In triangle ABC , m(^BAC)=120∘ . Let the bisectors of angles ˆA , ˆB and ˆC meet the opposite sides at D , E and F respectively. Prove that DE⊥DF . Proof I. Denote S∈BE , AS⊥AE and S′∈BE∩DF . Observe that the ray [AS is the bisector of ^IAB and AS⊥AE ⟹ the division (B,S,I,E) is harmonically. Is well-known that the division (A,S′,I,E) is harmonically. Thus, S′≡S , i.e. the ray [¯DSF is the bisector of ^ADB . Analogously prove that the ray [DE is the bisector of ^ADC . In conclusion, DF⊥DE . See and here . Proof II. ‖ \implies DE\perp DF . Proof III. If denote the intersection L\in CF\cap DE , then D(F,I,L,C) is harmonical pencil, i.e. (\infty , X,A,Y) is harmonical division \implies AX = AY = \frac {bc}{b + c} .If denote the second intersection A' of AI with the circumcircle of ABC , then BCA' is equilateral. Therefore, AA' = b + c (Pompeiu's theorem). But AA'\cdot AD = bc ( ABA'\sim ADC ). In conclusion, AD = \frac {bc}{b + c} . Otherwise, you can construct outside of ABC equilateral triangles a.s.o.
17.12.2021 23:48
Since \overline{AC} bisects the exterior angle \angle BAD, E must be the B-excenter of \triangle ABD. By symmetry, F must be the C-excenter of \triangle ACD. Thus, \overline{DE} bisects \angle ADC, while \overline{DF} bisects \angle ADB, from which we can find that \angle FDE = \tfrac{1}{2}\angle BDC = 90^\circ. The desired conclusion follows immediately.
02.04.2022 20:43
Notice \measuredangle (\overline{CA},\overline{AB})=60=\measuredangle DAC so E is the B-Excenter of \triangle ABD. Hence, \angle CDE=\angle EDA, and similarly \angle ADF=\angle FDB. Thus, 2\angle EDF=\angle BDC=180. \square