Let AL and BK be angle bisectors in the non-isosceles triangle ABC (L lies on the side BC, K lies on the side AC). The perpendicular bisector of BK intersects the line AL at point M. Point N lies on the line BK such that LN is parallel to MK. Prove that LN=NA.
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Tags: geometry, circumcircle, trigonometry, perpendicular bisector, angle bisector, geometry proposed
20.06.2010 20:53
I think it's nice choice for juniors. Here is my solution. Firstly, we check the case if AB=AK. Then AM coincides with perpendicular bisector of BK, so BK⊥AM. Moreover, because BK is bisector and altitude in △ABL, it's perpendicular bisector of AL. Thus, NL=NA. Now, AB≠AK. Let circumcircle of ABK meets AM at M′. Then KM′=M′B, meaning that, if M and M′ are distinct, MM′ is perpendicular bisector of BK ⟹AK=AB, impossible. Thus, M=M′ and quadrilateral BMKA is cyclic. Then ∠KBA=∠KMA=∠NLA Quadrilateral BLNA is also cyclic, and because BN is bisector, we finally get LN=NA.
21.06.2010 00:08
21.06.2010 17:10
Because BM=MK and ∠BAM=∠MAK ,after applying the Theorem of Sinuses in △ABM and in △MAK we see that sin∠ABM=sin∠AKM so we conclude that either ∠ABM=∠AKM or ∠ABM=180−∠AKM which will mean that BAKM is cyclic.Checking the first case we get ∠B2+∠IBM=∠B2+∠IBM+∠C which means ∠C=0 so contradiction ,which confirms that ABMK is cyclic.Moving further we get ∠BAM=∠BKM=∠BNL so we conclude that BANL is cyclic too.And the result comes by himself AN=sinB2⋅2R=AL where R is the radius of the cercle passing through points B,A,N and L. So we are done.
21.06.2010 23:16
Ahiles wrote: Let AL and BK be angle bisectors in the non-isosceles triangle ABC , where L∈(BC) and K∈(AC) . The perpendicular bisector of BK intersects the line AL at point M. Point N lies on the line BK such that LN is parallel to MK. Prove that LN=NA. Remark. An easy and nice proposed problem ! Observe that always AB>AK . Indeed, m(^AKB)=B2+C>B2= m(^ABK) ⟹ m(^AKB)>m(^ABK)⟹AB>AK . Otherwise, AB>AK⟺c>bca+c⟺a+c>b what is truly always ! Quote: An easy extension. Let AL be the angle bisector in the non-isosceles triangle ABC , where L∈(BC) and let X∈(AL) be a point for which denote K∈BX∩AC . The perpendicular bisector of BK intersects the line AL at the point M. The point N lies on the line BK such that LN∥MK. Prove that NANL=XAXL⋅ILIA . Proof (similarly with frenchy's or Ahiles'). Is well-known that "in any nonisosceles triangle the intersection point between the bisector of an angle of triangle and the perpendicular bisector of the opposite side belongs to the circumcircle of the given triangle". Apply this property to the triangle ABK and the vertex A . Thus obtain that the quadrilateral ABMK is cyclically ⟹ ^BAM≡^BKM≡^BNL⟹ ^BAL≡^BNL⟹ the quadrilateral BANL is cyclically ⟹ NANL=sin^NBAsin^NBL=sin^XBAsin^XBL . Since XAXL= BABL⋅sin^XBAsin^XBL= IAIL⋅sin^XBAsin^XBL obtain sin^XBAsin^XBL=XAXL⋅ILIA . In conclusion NANL=XAXL⋅ILIA . Remark. In the particular case when X:=I obtain the proposed problem. Another interesting particular case obtain such : denote P∈AB∩CI , R∈AC∩BI , L∈BC∩AI and S∈PR∩AI . Then for X:=S obtain NA=2⋅NL . This results from the well-known relation IAIL=2⋅SASL .
25.06.2010 15:34
Wow. I feel so stupid. It took me so long to see the major idea. Solution Since AL is the bisector of ∠KAB, the intersection of AL with the circumcircle of △AKB is the midpoint of minor arc KB. This point also lies on the perpendicular bisector of KB and therefore is M. Hence AKMB is cyclic and ∠ABK=∠AMK=∠ALN. Therefore ANLB is cyclic. Since BK bisects ∠ABL, N is the midpoint of minor arc AL and therefore LN=NA.
04.01.2011 20:04
See this: ABMK cyclic, hence ∠AMK=∠ABK, but ∠ABK=∠ABN and ∠AMK=∠ALN, consequently ABLN cyclic as well; with BN angle bisector, done. Best regards, sunken rock
28.10.2014 13:20
Dear Mathlinkers, for proving that ABLN is cyclic , we can use a converse of the Reim's theorem. Sincerely Jean-Louis
27.07.2016 14:02
can somebody please post a figure?
24.06.2019 22:56
A little late, but here's a diagram for anyone else wanting one. It includes the two important circumcircles. [asy][asy] unitsize(1.5cm); pair A, B, C, N, M, K, L; A=origin; B=(4.5,4); C=(7,0); L=extension(A, bisectorpoint(B,A,C), B, C); K=extension(B, bisectorpoint(A,B,C), A, C); M=extension(A, L, midpoint(B--K), bisectorpoint(B,K)); N=extension(B,K,L,L+M-K); draw(midpoint(B--K)--M); draw(A--B--C--cycle); draw(A--L); draw(L--N); draw(M--K); draw(B--N); draw(A--N); draw(circumcircle(A,B,K)); draw(circumcircle(A,B,N)); dot(M); label("A", A, SW); label("B", B, dir(60)); label("C", C, SE); label("L", L, NE); label("K", K, 2dir(310)); label("M", M, dir(320)); label("N", N, dir(300)); [/asy][/asy]
27.07.2021 05:11
Dropping altitudes from M gives congruent triangles that imply (BLMC). (BLMC) implies (BNKC). Therefore, ∠NBK=∠NKB
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01.01.2022 12:05
In triangle ABK, angle bisector of ∠A and perpendicular bisector of BK meet at M so ABKM is cyclic. ∠ABK = ∠KMA = ∠NLA ---> ABLN is cyclic. ∠LAN = ∠LBN = ∠NBA = ∠NLA ---> NL = NA.
25.06.2022 08:53
[asy][asy] import olympiad; unitsize(50); pair A, B, C, Q, P, R, S, M, O; // Coordinates A=(0,0); B=(3,0); C=(3.4,3.5); Q = extension(A, bisectorpoint(C,A,B), B, C); P = extension(B, bisectorpoint(C,B,A), A,C); R = extension(midpoint(B--P), bisectorpoint(B,P), A, Q); S = extension(B, P, Q, Q+R-P); M = foot(R,P,B); O = circumcenter(A,P,B); // Drawing draw(A--B); draw(B--Q); draw(Q--C); draw(A--P); draw(P--C); draw(A--Q); draw(B--S); draw(midpoint(B--P)--R); draw(R--P); draw(Q--S); draw(circumcircle(S,Q,B)); draw(circumcircle(A,P,R)); draw(B--R); draw(A--S); dot(O); draw(O--P, dashed); draw(O--B, dashed); draw(M--O); // Labels label("A", A, NW); label("B", B, NE); label("C", C, N); label("Q", Q, NE); label("P", P, N); label("R", R, N); label("S", S, NW); label("M", M, N*dir(30)); label("O", O, NE); [/asy][/asy] Claim: Quadrilaterals APRB and ASQB are cyclic. Proof: Let R′ be the point which bisects minor arc BP. We will show R′=R. We start with the angle bisector of ∠BAC. Note that equal angles subtend equal arcs, so the angle bisector of ∠BAC obviously intersects point R′. Now, let's consider the perpendicular bisector of BP. Let M be the midpoint of BP and O be the circumcenter of △APB. Notice ∠PMO=∠BMO=90∘ by the perpendicular bisector condition and OM=OM, so by RHS Congruence, △PMO≅△BMO. Thus, we know ∠POR=∠BOR, which implies arcs PR and BR are equal, as equal central angles are subtended by equal arcs. Therefore, we have shown OR bisects minor arc BP, (so R is the midpoint of minor arc BP) implying the perpendicular bisector of BP lies on R=R′, so APRB is cyclic. Now, with this information, let's take a look at quadrilateral ASQB. Recall that QS∥RP. So, ∠BAR=∠BPR=∠BSQ. Thus, ASQB is cyclic, and our claim is proven. Claim: Segments QS and SA are equal. Proof: Recall that BP is the angle bisector of ∠ABC, so ∠ABS=∠SBQ (Note S is simply an extension of BP). Furthermore, if two central angles in a circle are equal, the corresponding chords are also equal in length. Thus, we have QS=SA, as desired.
02.07.2022 20:23
@above, you need to redefine Q↦L,P↦K,M↦R,N↦S. Let AL intersect the circumcircle of △ABK at M′. Since ∠BAK is inscribed, angle bisector AL meets the circumcircle of △ABK at the midpoint of minor arc BK. Since M′ is the midpoint of minor arc BK, it must also be on the perpendicular bisector of BK. But since the perpendicular bisector of BK intersects AL at M, we must have M′=M. Therefore, quadrilateral BAKM is cyclic. Hence ∠ABK=∠AMK. But since KM∥LN, we must also have ∠ABK=∠AMK=∠ALN. So quadrilateral LBAN must also be cyclic. Since BN is a perpendicular bisector and ∠ABL is inscribed, it follows that N is the midpoint of \overarcAL. Thus, we have LN=NA as desired.
02.12.2023 21:26
Since M lies on the angle bisector of ∠ABK and on the perpendicular bisector of BK, we obtain that ABMK is cyclic (indeed AB≠AK, else ∠AKB=∠ABK=∠CBK, which is impossible since ∠AKB is an external angle in triangle CBK). Hence ∠ABN=∠ABK=∠AMK=∠ALN (the latter from MK∥LN), so ABLN is cyclic. Since BN bisects ∠ABL, we obtain AN=NL, as desired.
03.01.2024 16:05
Back to the problem. The diagram is given below. Since DM is the perpendicular bisector of BK, BM=MK. So, by our Lemma, ABMK is cyclic as ∠BAM=∠MAK. Notice that, BC∦ which implies \angle AKB\ne \angle KBC=\angle ABK\implies AB\ne AK. Since MK\parallel NL, \angle BNL=\angle BKM=\angle BAM=\angle BAL\implies ABLN\text{ is cyclic}. Therefore, \angle LAN=\angle LBN=\angle ABN=\angle ALN\implies AN=LNand we are done.
31.01.2025 06:40
Redefine N as intersection of perpendicular bisector of AL with BK, then clearly from the definitions and angle bisector of midpoints of arcs (alternatively just ghost point) we get that ANLB, AKMB are cyclic and now Reim's just finishes.