Given is an acute angled triangle $ \triangle ABC$ with side lengths $ a$, $ b$ and $ c$ (in an usual way) and circumcenter $ O$. Angle bisector of angle $ \angle BAC$ intersects circumcircle at points $ A$ and $ A_{1}$. Let $ D$ be projection of point $ A_{1}$ onto line $ AB$, $ L$ and $ M$ be midpoints of $ AC$ and $ AB$ , respectively. (i) Prove that $ AD=\frac{1}{2}(b+c)$ (ii) If triangle $ \triangle ABC$ is an acute angled prove that $ A_{1}D=OM+OL$
Problem
Source: Federation of Bosnia and Heryegovina, 2nd grades, 2008.
Tags: geometry, circumcircle, rectangle, trigonometry, angle bisector
The QuattoMaster 6000
29.04.2008 03:24
delegat wrote: Given is an acute angled triangle $ \triangle ABC$ with side lengths $ a$, $ b$ and $ c$ (in an usual way) and circumcenter $ O$. Angle bisector of angle $ \angle BAC$ intersects circumcircle at points $ A$ and $ A_{1}$. Let $ D$ be projection of point $ A_{1}$ onto line $ AB$, $ L$ and $ M$ be midpoints of $ AC$ and $ AB$ , respectively. (i) Prove that $ AD = \frac {1}{2}(b + c)$ (ii) If triangle $ \triangle ABC$ is an acute angled prove that $ A_{1}D = OM + OL$
Consider point $ P$ on $ AB$ such that $ AD = PD$ and $ AP = 2AD$. It follows that $ A_1D$ is a perpindicular bisector of $ AP$, so $ \angle A_1AC = \angle PAA_1 = \angle DPA_1$. Also, $ \angle A_1CA = \angle A_1BP$ and $ \angle A_1CB = \angle A_1AB = \angle A_1AC = \angle A_1BC$, so $ A_1B = A_1C$. It follows that $ \triangle A_1BP\cong \triangle A_1CA\implies BP = AC$. Thus, $ 2AD = AP = AB + BP = AB + AC\implies AD = \frac {b + c}{2}$.
Extend $ MO$ to $ Q$ such that $ OQ = OL$. Notice that $ \angle A_1OQ = \angle MOA = 90 - \angle OAM = 90 - \angle OAL = \angle LOA$. Also, $ OA_1 = OA$. Thus, $ \triangle OLA\cong \triangle OQA_1$, so $ \angle OLA = \angle OQA_1 = \angle MQA_1$. Yet, $ O$ is the circumcenter of $ \triangle ABC$, so $ \angle QMD=\angle OMD = 90$ and $ \angle OLA = 90$, so $ \angle MQA_1=\angle OQA_1 = 90$. Since $ \angle A_1DM=90$, we have that $ A_1DMQ$ is a rectangle, so $ A_1D = MQ = OM + OQ = OM +OL$.
Virgil Nicula
29.04.2008 07:55
delegat wrote: Given is an acute angled triangle $ \triangle ABC$ with side lengths $ a$, $ b$ and $ c$ (in an usual way) and circumcenter $ O$. Angle bisector of angle $ \angle BAC$ intersects circumcircle at points $ A$ and $ A_{1}$. Let $ D$ be projection of point $ A_{1}$ onto line $ AB$, $ L$ and $ M$ be midpoints of $ AC$ and $ AB$ , respectively. Prove that $ AD = \frac {b + c}{2}$ and $ A_{1}D = OM + OL$ . Proof. Observe that $ AA_1 = 2R\cdot\sin (B + \frac A2)$ and $ \{\begin{array}{ccc} b=2R\sin B & , & c=2R\sin C\\\\ OM = R\cos C & , & OL = R\cos B\end{array}$ . $ \blacktriangleright$ $ AD = AA_1\cdot\cos\frac A2 = 2R\sin(B + \frac A2)\cos\frac A2 = R(\sin B + \sin C) =\frac {b+c}{2}$ . $ \blacktriangleright$ $ A_1D = AA_1\cdot\sin\frac A2 = 2R\sin(B + \frac A2)\sin\frac A2 = R(\cos B + \cos C) = OL + OM$ .
ZyXel
30.07.2008 14:27
I have nice proof for AD=(b+c)/2 . My proof for A1D=OL+OM is same as Virgil`s. Without loss of generality we can assume that AC<AB. Consider point K on AB such that AC=AK and point E as intersection of AA1 and CK. It`s obvious that CE=KE and CA1= A1K. Since ABA1C is cyclic, BA1= A1C= A1K. So, triangle A1KB is isosceles and KD=BD. 2AD=2AK+2KD=AC+AK+KD+DB=AC+AB and AD=(b+c)/2
nayel
30.07.2008 18:31
WLOG assume that $ AB\le AC$ and let $ C'$ be the point on ray $ AB$ such that $ AC=AC'$. Since $ A_1$ is the midpoint of arc $ BC$ we get $ A_1B=A_1C=A_1C'$ implying $ \triangle A_1BC'$ is isosceles. Therefore $ A_1D\perp BC'\implies D$ is the midpoint of $ BC'$ and thus $ AD=\frac{AB+AC'}{2}=\frac{AB+AC}{2}$.
arshakus
29.05.2011 20:02
1)$2AD=b+c<=> MC=BD->$ because $\triangle ADA_1=\triangle AA_1M$, from $\triangle BDA_1=\triangle A_1MC$ we will get $MC=BD=> PROOVED$