Midpoint of base in isosceles triangle [<APM + <BPC = 180°]

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warut_suk

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warut_suk

#1 h Jan 25, 2005, 11:10 AM • 3 Y

Y by Adventure10, jhu08, Mango247

Let a triangle $ABC$ satisfy $AC = BC$ ; in other words, let $ABC$ be an isosceles triangle with base $AB$ . Let $P$ be a point inside the triangle $ABC$ such that $\angle PAB = \angle PBC$ . Denote by $M$ the midpoint of the segment $AB$ . Show that $\angle APM + \angle BPC = 180^{\circ}$ .

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grobber

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grobber

#2 h Jan 25, 2005, 11:17 AM • 5 Y

Y by Adventure10, jhu08, Mango247, and 2 other users

By an angle chase, we can see that $P$ moves on a circle tangent in $A,B$ to $CA,CB$ respectively. Let $Q$ be the second intersection of $CP$ with this circle. $PAQB$ is a harmonic cyclic quadrilateral, so, if $C'$ is the intersection between the tangent at $P$ to the circle and $AB$ , we find $(PA,PB;PQ,PC')=-1$ , meaning that $PQ$ is the symmedian of $PAB$ , which is precisely what we want.

I was in a hurry while writing this, but I'll clarify if needed.

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darij grinberg

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darij grinberg

#3 h Jan 25, 2005, 11:23 AM • 3 Y

Y by Adventure10, jhu08, Mango247

Rewriting your problem with different notations:

Let ABC be an isosceles triangle with AB = AC. Let P be a point inside the triangle ABC such that < PBC = < PCA. Let M be the midpoint of the segment BC. Prove that < BPM + < APC = 180°.

Now, this problem was solved in posts #8 and #9 of http://www.mathlinks.ro/Forum/viewtopic.php?t=18258 (see the Lemma in post #8), with the only difference being that an additional assumption was used, namely the assumption that < PCB = < PBA; but this assumption is actually unnecessary, since it follows from the other assumptions of the problem (namely, since the triangle ABC is isosceles with AB = AC, we have < ACB = < ABC, and since < PBC = < PCA, we have < PCB = < ACB - < PCA = < ABC - < PBC = < PBA, so that < PCB = < PBA comes out as a consequence of the other assumptions of the problem).

Darij

This post has been edited 2 times. Last edited by darij grinberg, Mar 15, 2007, 9:35 AM

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Arne

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Arne

#4 h Aug 22, 2005, 10:39 AM • 7 Y

Y by LoveMath4ever, r31415, Adventure10, jhu08, Mango247, and 2 other users

warut_suk wrote:

Let a triangle $ABC$ satisfy $AC = BC$ ; in other words, let $ABC$ be an isosceles triangle with base $AB$ . Let $P$ be a point inside the triangle $ABC$ such that $\angle PAB = \angle PBC$ . Denote by $M$ the midpoint of the segment $AB$ . Show that $\angle APM + \angle BPC = 180^{\circ}$ .

Take point Q such that APBQ is a parallellogram.

Take point R on AQ such that ACB, APR are similar. (I assume R lies between A and Q, the other case is similar) Then BPRQ is cyclic, and ACP, ABR are similar.

Hence BPQ = BRQ = 180 - ARB = 180 - APC. Hence result.

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Virgil Nicula

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Virgil Nicula

#5 h Jan 15, 2006, 3:19 PM • 7 Y

Y by LoveMath4ever, Abdollahpour, Adventure10, jhu08, Mango247, and 2 other users

$S\in AB\cap CP$ , $CA=CB=a$ , $AB=b$ , $m(\widehat {PCA})=u$ , $m(\widehat {PCB})=v$ ,
$m(\widehat {PAB})=m(\widehat {PBC})=x$ , $m(\widehat {PAC})=m(\widehat {PBA})=y$ .

$\blacktriangle$ The theorem of Sinus in the triangle $APB\Longrightarrow \frac{PA}{PB}=\frac{\sin y}{\sin x}\ \ (1)$ .
$\blacktriangle$ The trigonometrical form of the Menelaus' theorem in
the triangle $ABC$ for the inner point $P\Longrightarrow$
$\sin u\sin^2x=\sin v\sin^2y$ $\Longrightarrow$ $\frac{\sin u}{\sin v}=\left(\frac{\sin y}{\sin x}\right)^2\ \ (2)$ .
$\blacktriangle\ \frac{SA}{SB}=\frac{CA}{CB}\cdot \frac{\sin \widehat {ACS}}{\sin \widehat {BCS}}$ $\Longrightarrow$ $\frac{SB}{SC}=\frac{\sin u}{\sin v}\ \ (3).$

From the relations $(1)$ , $(2)$ , $(3)$ results: $\frac{SA}{SB}=\left(\frac{PA}{PB}\right)^2$ , i.e. the ray $[PS$ is
the symmedian from the vertex $P$ in the triangle $APB$ , meaning

$\widehat {APM}\equiv \widehat {BPS}\Longrightarrow$ $\boxed {\ m(\widehat {APM})+m(\widehat {BPC})=180^{\circ}\ }.$

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treegoner

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treegoner

#6 h Jan 15, 2006, 5:05 PM • 2 Y

Y by Adventure10, jhu08

The problem can be generalized as follow (which is well-known)
Let $PAB$ be a triangle and $M$ be the midpoint of $AB$ . Suppose $C$ is the intersection of the tangents of the circumcircle at $A$ and $B$ . Then $\measuredangle{BPM}= \measuredangle{CPB}$ .
[Moderator edit: This is basically the assertion of http://www.mathlinks.ro/Forum/viewtopic.php?t=99571 .]
Back to the problem , it is easy to see that $CA$ and $CB$ are the tangents of $(PAB)$ at $A$ and $B$ .

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Virgil Nicula

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Virgil Nicula

#7 h Jan 15, 2006, 6:31 PM • 3 Y

Y by Adventure10, jhu08, Mango247

See http://www.mathlinks.ro/Forum/viewtopic.php?t=55581
http://www.mathlinks.ro/Forum/viewtopic.php?t=43202
where are mentioned the properties of the harmonical quadrilateral.

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zbghj

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zbghj

#8 h Oct 13, 2007, 5:49 PM • 3 Y

Y by Adventure10, jhu08, Mango247

Let us see the circumcircle of APB. Let v=<ABC.
Use the center of the circumcircle as origin of coordinate and give the coordinates as:

C (0,1)
M (0, (cos v)^2)
P (cos v*cos u, cos v*sin u)

It is easy to obtain: PM/PC=cos v

So, we can get: BC/PC=AM/PM.
So, <BPC=<APM or <BPC+<APM=180.

When <BPC=<APM, then <AMP=<BCP. So, <BPC=<BMC=90. As a result, <BPC+<APM=180 too.

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yetti

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yetti

#9 h Dec 8, 2007, 2:13 AM • 3 Y

Y by Adventure10, jhu08, Mango247

P is on circle (U) tangent to BC at B and passing through A(then $\angle PAB = \angle PBC$ ). $\triangle ABC$ is isosceles $\Longrightarrow$ (U) is also tangent to CA at A, AB is polar of C WRT the circle (U). CP meets (U) at P and again at Q. If P' is a reflection of P in $CM \perp AB,$ $P' \in (U)$ and CP' meets (U) again at Q'. PP'Q'Q is isosceles trapezod symmetrical WRT CM, the diagonals PQ', P'Q meet on CM and on the polar AB of C, $M \equiv PQ' \cap P'Q \Longrightarrow$ $\angle APM \equiv \angle APQ' = \angle BP'Q = \angle BPQ = 180^\circ - \angle BPC.$

This post has been edited 1 time. Last edited by yetti, Dec 9, 2007, 3:02 AM

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Leonhard Euler

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Leonhard Euler

#10 h Dec 12, 2007, 8:44 AM • 3 Y

Y by Adventure10, jhu08, Mango247

We have $\angle PBA=\angle PAC$ . Let $X$ be on $PM$ such $PAXB$ is parallogram. Then $\angle XAB=\angle PAC,\angle XBA=\angle PBC$ . Hence $C,X$ are isogonal points wrt $\triangle PAB$ . Hence $PC,PX$ is isogonal line $\triangle PAB$ . This imply $\angle APM+\angle BPC=180$ .

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Learner94

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Learner94

#11 h Dec 22, 2011, 12:19 PM • 4 Y

Y by samirka259, jhu08, Adventure10, Mango247

Lemma: Let $ABC$ be a triangle and $\Gamma$ its circumcircle. Let the tangent to $\Gamma$ at $B$ and $C$ intersect at $D$ . Then $AD$ coincides with the $A$ - symmedian of $\triangle ABC$ . The proof can be found here.

By angle chasing we can show that $AC$ and $BC$ are tangents to the circumcircle of $\triangle PAB$ at points $A$ and $B$ respectively. The tangents $AC$ and $BC$ intersect at $C$ . So $PC$ coincides with the $C$ - symmedian of $\triangle ABC$ . Since $PM$ is the median of $\triangle ABC$ , $\angle APM = \angle BPE$ . or $\angle APM = 180^{\circ} - \angle BPC$ . So $\angle APM + \angle BPC = 180^{\circ}$ .

Attachments:

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DaChickenInc

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DaChickenInc

#12 h May 18, 2014, 5:47 PM • 3 Y

Y by jhu08, Adventure10, Mango247

Translation of Official Solution
Official Solutions
Problem Significance

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Shauryajain123

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Shauryajain123

#13 h Mar 4, 2021, 4:51 AM • 2 Y

Y by AllanTian, jhu08

A nice problem for symmedians

We see that AC and AB are tangents to the circumcircle of $\triangle{APB} \implies CP$ is the symmedian.
Then the rest easily follows from how symmedians are defined

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PCChess

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PCChess

#14 h Mar 12, 2021, 5:55 PM • 1 Y

Y by jhu08

Since $\triangle ABC$ is isosceles, we know that $\angle PAC=\angle PBA$ . Consider $(APB)$ . Since $\angle PAB=\angle PBC$ and $\angle PAC=\angle PBA$ , lines $AC$ and $BC$ are tangent to $(APB)$ . This implies that $PC$ lies on the $P$ -symmedian. Let $CP$ intersect $AB$ at point $E$ . By the definition of the symmedian, $\angle APM=\angle BPE$ . Clearly $\angle BPE+\angle BPC=180^{\circ}$ , so $\angle APM + \angle BPC = 180^{\circ}$ .

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JustKeepRunning

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JustKeepRunning

#15 h Jun 9, 2021, 12:13 AM • 1 Y

Y by jhu08

Let $CP\cap AB=G$ . Then the problem is equivalent to $\angle APM=\angle BPG$ , aka. $CG$ is a symmedian in $\triangle APG$ . Notice that $CB$ and $CA$ are the tangents to $(APB),$ so we are done.

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Mogmog8

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Mogmog8

#16 h Dec 11, 2021, 1:10 AM • 2 Y

Y by centslordm, jhu08

Since $\angle CBP=\angle BAP$ and $\angle PAC=\angle PBA,$ $C$ is the intersection of the tangents to $(PAB)$ at $A$ and $B.$ Hence, $\overline{PC}$ is the $P$ -symmedian point of triangle $PAB.$ Notice that $\measuredangle BPC=-\measuredangle(\overline{CP},\overline{PB})=-\angle APM.$ $\square$

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JustKeepRunning

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JustKeepRunning

#17 h Dec 11, 2021, 1:48 AM • 1 Y

Y by jhu08

Nice problem!

It is well known that $P$ is the $A$ humpty point, and so $M_1:=AP\cap BC$ is the midpoint of $BC$ . Furthermore, we have that $\angle PCB=\angle PAC=\angle PBA$ and $\angle PAB=\angle PBC$ . Hence, we get that $\angle APM+\angle BPC=\angle APM+\angle CPM_1+\angle M_1PB=\angle APM+\angle MPB+\angle BPM_1=180^{\circ},$ where this follows from similar triangles $\triangle PMB$ and $\triangle PM_1C$ .

EDIT: oops just realized I already solved this before. At least this is a slightly different/more motivated solution.

This post has been edited 1 time. Last edited by JustKeepRunning, Dec 11, 2021, 1:48 AM

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Albert123

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Albert123

#18 h Dec 11, 2021, 5:13 AM • 1 Y

Y by jhu08

Let $(APB) \cap CP=L$
As well: $CA$ and $CB$ are tangent of $(APB)$
Then: $PQ$ is symedian of $(APB)$
$\implies \angle APM=\angle LPB \implies \angle APM + \angle BPC=180$

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#19 h Feb 16, 2022, 1:20 PM

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Let $CP$ meet circumcircle of $APB$ at S. Since $\angle PAB = \angle PBC$ then $CB$ is tangent to $APB$ and since $CB = CA$ then $SP$ is symmedian of $ASB$ so $\angle APM = \angle BPS$ and so $\angle APM + \angle BPC = \angle 180$ .

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EulersTurban

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EulersTurban

#20 h Feb 16, 2022, 5:54 PM

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Let $CP \cap AB = D$ . The angle condition implies that $AC$ and $AB$ are tangent to the circumcircle of $APB$ , then we know that $PD$ is the $P$ -symmedian of $APB$ , which implies that $\angle TPC = \angle APD = \angle MPB$ thus we have that $\angle BPC = \angle MPT$ , which clearly gives us that $\angle APM + \angle BPC = 180$

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HamstPan38825

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HamstPan38825

#21 h Jul 31, 2023, 3:10 PM

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Let $\overline{PM'}$ be a symmedian in triangle $BPC$ , so it suffices to show that $A$ lies along the symmedian. But this is evident by the tangent intersection definition.

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Ianis

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Ianis

#22 h Jul 31, 2023, 6:41 PM

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Complex

Synthetic

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Grasshopper-

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Grasshopper-

#23 h Jul 26, 2024, 10:52 AM

Y by

Let CP intersect AB at N, the circumradius of tri. ACP(center O) and tri. BCP(center O') be r, r' respectively. We have to prove that CP is P -symmedian in tri. APB and to show that we need to prove AN/BN=AP^2/BP^2 ,
We have [tri. APN]/[tri. BPN] = AN/BN = [tri. APC]/[tri. BPC]=(AP × sin<CAP)/(BP × sin<CBP)
by sine rule,
sin<CAP/sin<CBP=r'/r, by angle chasing we will get that tri. OO'P and tri. APB are similar so, r'/r = AP/BP then, we get that AN/BN=AP^2/BP^2
so, PN is P-symmedian in tri. APB. This implies <APM = <BPN; <BPN + <CPB=180°=> <APM + <BPC=180° .

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