Midpoint of angle bisector

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Midpoint of angle bisector

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#1 h Jun 27, 2014, 5:22 PM • 5 Y

Y by Davi-8191, Mathotsav, Adventure10, Mango247, and 1 other user

Let $M$ be midpoint of angle bisector $AD$ of triangle $ABC$ . Circle $k_1$ with diameter $AC$ cuts $BM$ at $E$ , and circle $k_2$ with diameter $AB$ cuts $CM$ at $F$ .Prove that $B$ , $E$ , $F$ and $C$ are concyclic.

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#2 h Jun 28, 2014, 3:01 PM • 4 Y

Y by frill, nikolapavlovic, Adventure10, Mango247

First think of the harmonic division, since it is nearly impossible to deal with $M$ otherwise. Let the parallel to $AD$ intersect $k_1$ at $P$ . Since $\angle CPA = 90 \implies AP$ is an external angle bisector. Therefore, $(PA, PD, PB, PC) = -1$ so $M \in PB$ . So, $\angle BEC = 90+A/2$ or $IBCE$ are concyclic where $I$ is the incentre of $ABC$ . Indeed, by symmetry, $F$ also lies on the circle.

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jayme

#3 h Jun 28, 2014, 3:28 PM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
I have also this reference
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=535799

Sincerely
Jean-Louis

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#5 h Jun 28, 2014, 4:32 PM • 2 Y

Y by Adventure10, Mango247

Here is an outline of my brute force solution:
We have that $AD^2=bc\left(1- \left (\frac{a}{b+c}\right)^2 \right)$
And we can get $BM$ as it is median inside $ABD$ , we can get $BD$ in Stewarts theorem,and $MB$ we can get using Stewarts theorem in triangles $AMB$ and $BMC$ and Pythagoras theorem in triangle $CEA$ . Analgously, we get $CM$ and $MF$ . Now this yields $ME^2 \cdot MB^2=MF^2 \cdot MC^2$ which yields $ME \cdot MB=MF \cdot MC$ which implies $B$ , $E$ , $F$ and $C$ are concyclic.

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#6 h Oct 10, 2014, 4:07 PM • 7 Y

Y by anantmudgal09, enhanced, Siddharth03, N1RAV, Adventure10, Mango247, and 1 other user

My solution:

Let $X, Y$ be the midpoint of $AC, AB$ , respectively .
Let $X', Y'$ be the intersection of $AD$ with $\odot (AC), \odot (AB)$ , respectively .

Easy to see $X, M, Y$ are collinear .

Since $\angle XAX' =\angle YAY'$ and $XA=XX',YA=YY'$ ,
so $AMX'X \sim Y'MAY$ . i.e. $MA \cdot MA=MX' \cdot MY'$
hence $\odot (AB) \longleftrightarrow \odot (AC) , B \longleftrightarrow E , C \longleftrightarrow F$ under Inversion $\mathbf{I}( \odot (AD) )$ ,
so $B, C, E, F$ are concyclic .

Q.E.D

This post has been edited 2 times. Last edited by TelvCohl, Apr 22, 2015, 4:15 PM

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#7 h Oct 31, 2014, 3:25 PM • 3 Y

Y by legendofmath, Adventure10, Mango247

I have a really nice solution to this great problem.
Obviously $\angle AEC=\angle AFB=90$ .
Now extend $BF$ by segment $FN$ such that $BF=FN$ .Triangle $ABN$ is isosceles so $\angle A-\angle CAF=\angle CAF+\angle CAN\Rightarrow 2(A/2-\angle CAF)=\angle CAN\Rightarrow \angle CAN=2\angle DAF$
Let $AK$ be the internal bisector of $\angle CAN$ where $K$ is on $CN$ .We have $\frac{KC}{KN}=\frac{AC}{AN}=\frac{DC}{DB}\Rightarrow KD\parallel BN$
But from Thales theorem and since we also know that $BF=FN$ we conclude that $L$ is the midpoint of $DK$ and so $ML\parallel AK$ .By angle chasing now we have $\angle DAF=\angle CAK=\angle ACF$ which means that $MA$ is tangent to the circumcircle of triangle $FAC$ and hence we obtain $MA^2=MF \cdot MC$ .
By following the same procedure (i.e. by extending $CE$ etc) we get that $MA^2=ME\cdot MB$ and so $ME\cdot MB=MF\cdot MC$ so $FEBC$ is cyclic.We also get that $MD^2=MA^2=MF \cdot MC$ which gives that $MD$ is tangent to the circumcircle of $BED$ and so $\angle EDM=\angle MBD=\angle MFB$ which finally gives the desired result.

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Radar

#8 h Nov 5, 2014, 10:29 PM • 3 Y

Y by tapir1729, Adventure10, and 1 other user

Just think of foot of altitude from $A$ to $BC$ (lets call it $T$ ) and then just invert the whole picture through $A$ . The result will be very symmetric and it will be suddenly all super-obvious (see the attachment).

Attachments:

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#9 h Apr 22, 2015, 4:01 PM • 1 Y

Y by Adventure10

Let the line through $B$ parallel to $AD$ intersect external bisector of $\angle A$ (the bisector which cuts $BC$ nearer to $B$ ) at point $X$ . By easy lenght chasing we get that $CM$ contains $X$ (we use Ceva theorem).Now as $\angle BXA=\angle BFA=90$ we get that $AXBF$ is cyclic so $\angle AFM=\angle AFX=\angle ABX=\frac{\angle A}{2}$ which implies that $MA$ is tangent to the circumcircle of triangle $\triangle AFC$ which implies $MF \cdot MC= MA^2$ . Analogously we get that $ME \cdot MB=MA^2$ so $BEFC$ is cyclic.

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#10 h Aug 26, 2015, 6:32 AM • 1 Y

Y by Adventure10

Let $BM, FM$ cut $k_1$ for a second time at $B', F'$ , respectively, and let $r_1, r_2$ be the radii of $k_1, k_2$ , respectively. Let $O_1, O_2$ be the midpoints of $\overline{AC}, \overline{AB}$ , respectively (i.e. the centers of $k_1, k_2$ ). Note that $M$ is the midpoint of $\overline{AD}$ which implies that $M \in O_1O_2.$ Meanwhile, $\tfrac{MO_1}{MO_2} = \tfrac{AO_1}{AO_2} = \tfrac{r_1}{r_2}$ by the Angle-Bisector Theorem. Therefore, $M$ is the internal center of homothety that swaps $k_1, k_2.$ Since $B', F'$ are the images of $B, F$ under this homothety, we find that $B'F' \parallel BF.$ Then note that $\tfrac{MF}{MB} = \tfrac{MF'}{MB'} = \tfrac{ME}{MC}$ because $C, E, B', F'$ are concyclic. It follows that $B, C, E, F$ are concyclic as well. $\square$

This post has been edited 1 time. Last edited by Dukejukem, Aug 26, 2015, 6:35 AM

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MouN

#11 h Oct 21, 2015, 7:04 PM • 2 Y

Y by Adventure10, Mango247

Let $a=BC$ , $b=CA$ , $c=AB$ and $d=AD$ . Furthermore let $N$ be the midpoint of $B$ and $C$ , and let $K$ be the foot of the altitude from $A$ on $a$ . Since $P=d\cap k_2$ is the foot of the altitude from $B$ to $d$ , and $d$ is the bisector of $b$ and $c$ , the homothety centered at $B$ with factor $2$ sends $P$ to a point on $b$ . Thus $P$ lies on the midline $x$ of $\triangle ABC$ parallel to $b$ . Analogously $Q=d\cap k_1$ lies on the midline $y$ parallel to $c$ . Note that $K$ lies on both $k_1$ and $k_2$ so $\angle (d,y)=\angle(d,c)=\angle(KP,a)$ and $PNQK$ is inscribed in a circle $\omega$ . Since $M$ is the circumcenter of $\triangle ADK$ and $x$ and $y$ pass through the centers of $k_2$ and $k_1$ we have $\angle(KQ,y)=90+\angle(AK,d)=\angle(MK,a)$ hence $MK$ is tangent to $\omega$ at $K$ , so the inversion centered at $M$ with power $AM^2$ fixes $A$ and $K$ and swaps $P$ and $Q$ , so it swaps $k_1$ and $k_2$ , hence it swaps $B$ with $E$ , and $C$ with $F$ , so $BEFC$ is cyclic.

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#12 h Dec 13, 2016, 8:09 PM • 2 Y

Y by Adventure10, Mango247

Let $I$ be the incenter and $I_A$ be the $A$ excenter of $ABC$ . Define $E',F'$ as the points on segments $BM$ and $CM$ , respectively such that $ME'\cdot MB=MD^2=MF'\cdot MC.$ Note that $(A,D;I,I_A)=-1 \Longrightarrow MD^2=MI\cdot MI_A,$ so we conclude that the six points, $B, E, I, F, C, I_A$ are concyclic. Finally, we have $\angle AE'B+\angle BE'C=\left(180^{\circ}-\frac{1}{2}\cdot \angle A \right)+\left(90^{\circ}+\frac{1}{2}\cdot \angle A \right)=270^{\circ} \Longrightarrow \angle AE'C=90^{\circ},$ from where we conclude $E'=E$ . Similarly, $F'=F$ and the result follows. $\square$

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#13 h Sep 29, 2017, 3:00 AM • 2 Y

Y by Adventure10, Mango247

Yet another solution

gobathegreat wrote:

Let $M$ be midpoint of angle bisector $AD$ of triangle $ABC$ . Circle $k_1$ with diameter $AC$ cuts $BM$ at $E$ , and circle $k_2$ with diameter $AB$ cuts $CM$ at $F$ .Prove that $B$ , $E$ , $F$ and $C$ are concyclic.

Redefine $E,F$ as HM-points in $\triangle ABD$ and $\triangle ACD$ . Drop perpendiculars $\overline{BX}, \overline{CY}$ on line $\overline{AD}$ .

Claim. $AFXB$ and $AEYC$ are cyclic.

(Proof) Observe that $(\overline{XY}, \overline{DA})=-1$ . Hence $\overline{XF} \perp \overline{DC}$ , so $\angle AFX=90^{\circ}+\angle CAM=180^{\circ}-\angle ABX$ proving $AFXB$ cyclic. Likewise, we see $AEYC$ is cyclic too. $\blacksquare$

Now it is amply clear $E, F$ are on $\odot(AC), \odot(AB)$ , respectively.

Remark. Both solutions here require the somewhat "gutsy" claim regarding $E,F$ being HM-points. It turns out that we exploit their properties in both solutions. Although here we don't make ad-hoc constructions and $\odot(AC), \odot(AB)$ appear more naturally.

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RC.

#16 h Oct 2, 2017, 4:49 AM • 3 Y

Y by themathfreak, Adventure10, Mango247

gobathegreat wrote:

Let $M$ be midpoint of angle bisector $AD$ of triangle $ABC$ . Circle $k_1$ with diameter $AC$ cuts $BM$ at $E$ , and circle $k_2$ with diameter $AB$ cuts $CM$ at $F$ .Prove that $B$ , $E$ , $F$ and $C$ are concyclic.

Using here the properties of HM-points also known as Humpty points by some people.

Lemma :: Let in $\Delta ABC$ ; $D$ be a point on extension of side $AB$ such that $\angle ACB = \angle DCB$ and $AM$ be the median then the circle with diameter $CD$ intersects $AM$ at $H_A$ which is the HM-point with respect to vertex $A$ .

Proof:

Back to the given problem Thus, $E$ is the HM-point of $\Delta BDA$ with respect to vertex $B$ . Thus, we have $BM * ME = AM^{2}$ and similarly, in $\Delta CDA$ , we have $CM * FM = AM^{2}$ . Thus $BM *ME = CM * FM$ and thus $BCFE$ is cyclic.

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#17 h May 1, 2019, 6:16 AM • 2 Y

Y by Adventure10, Mango247

Here's another solution: Let $X, Y$ be points on $AB, AC$ such that $CX \parallel BY \parallel AD$ . Also let $E'$ and $F'$ be the midpoints of segments $CX$ and $BY$ respectively. Finally denote the foot of the $A$ -altitude by $H$ . Then we get $\angle CXA=\angle DAB=\angle DAC=\angle ACX \Rightarrow AC=AX \Rightarrow AE' \perp CX$ This means that $E' \in \odot (AHC)$ . Also, by homothety at $B$ , we get that $E' \in BM$ , or equivalently that $B,E,E'$ are collinear. As $CE' \parallel MD$ , so by converse of Reim's Theorem, we have $E \in \odot (DHM)$ . Similarly $F$ also lie on this circle, and so $E, F, H, D, M$ are concyclic.

Now, $\angle AHD=90^{\circ}$ , which means that $M$ is the circumcenter of $\triangle AHD$ . Considering the power of $C$ wrt $\odot (AHD)$ , we get that $CD \cdot CH=CF \cdot CM$ , which is equivalent to saying that $F$ is the inverse of $C$ in $\odot (AHD)$ . Thus, $MF \cdot MC=MD^2$ , and so $MD$ is tangent to $\odot (CFD)$ . This gives $\angle MDF=\angle DCF \Rightarrow \angle MEF=\angle MCB \Rightarrow B, C, E, F \text{ are concyclic. } \blacksquare$

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Promi

#18 h Jul 30, 2019, 5:32 PM • 2 Y

Y by Adventure10, Mango247

Let $N$ be the intersection of $CM$ and the perpendicular line through $A$ on $AM$
Define $J=CN\cap AB$
So, $(N,M;J,C)=-1$
So, taking the pencil from point $B$ we get, $BN||AM$
$N$ lies on the circle with diameter $AB$
Consequently, $BN||AM||CH$ , where $H=NA\cap BM$
$H$ lies on the circle with diameter $CH$
Let $K$ be the second intersection of the two circles, which lies on $BC$
Then, $\angle{KFM}=\angle{KAN}$
And $\angle{KEM}=\angle{KAH}$
So, $\angle{KEM}=\angle{KFM}$
So, $KEMF$ is cyclic.
Then simple angle chasing gives $\triangle MED$ and $\triangle MKB$ similar.
So, $MK^2=ME.MB$
Similarly, $MK^2=MF.MC$
We are done.

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