Reflection and tranformations

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

arandomperson123

430 posts

arandomperson123

#1 h Feb 17, 2017, 4:14 PM • 2 Y

Y by Adventure10, Mango247

Consider a triangle $AMC$ with $AB = AC$ and points $M, N$ lie on $AB$ , $AC$ , respectively. The lines $BN$ and $CM$ intersect at P. Prove that $MN$ and $BC$ are parallel iff $\angle{APM} = \angle{APN}$ .

Please solve using geometric transformations.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

nikolapavlovic

1246 posts

nikolapavlovic

#2 h Feb 17, 2017, 4:21 PM • 1 Y

Y by Adventure10

In the first direction is just $\triangle APM\cong APN$ so the desired.In the second direction we have that $AP$ is the median in $\triangle ABC$ and thus it's well-known that $MN||BC$ (for example by $(B,C;\text{midpoint of BC},\infty)=-1$ ).

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

arandomperson123

430 posts

arandomperson123

#3 h Feb 17, 2017, 4:40 PM • 2 Y

Y by Adventure10, Mango247

nikolapavlovic wrote:

In the first direction is just $\triangle APM\cong APN$ so the desired.In the second direction we have that $AP$ is the median in $\triangle ABC$ and thus it's well-known that $MN||BC$ (for example by $(B,C;\text{midpoint of BC},\infty)=-1$ ).

Why is the median of $ABC$ being $AP$ ?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

nikolapavlovic

1246 posts

nikolapavlovic

#4 h Feb 17, 2017, 4:55 PM • 2 Y

Y by arandomperson123, Adventure10

Wow,sorry points got mixed up.Here's a new proof:
If $AP\perp BC$ the conclusion is immediate so suppose not. $\angle BPC<\angle ABC,\angle PCB<\angle ACB$ $\implies$ $\angle BPC>\angle BAC$ .Now the angle bisector of $\angle BPC$ and the perp bisector of $BC$ intersect in $A$ but it's well-known the intersection is on the circle $\odot BPC$ but $A,P$ are on the same side of $BC$ and $\angle BPC\not =\angle A$ a contradiction hence the conclusion.

This post has been edited 2 times. Last edited by nikolapavlovic, Feb 17, 2017, 4:56 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

arandomperson123

430 posts

arandomperson123

#5 h Feb 17, 2017, 5:35 PM • 1 Y

Y by Adventure10

nikolapavlovic wrote:

Wow,sorry points got mixed up.Here's a new proof:
If $AP\perp BC$ the conclusion is immediate so suppose not. $\angle BPC<\angle ABC,\angle PCB<\angle ACB$ $\implies$ $\angle BPC>\angle BAC$ .Now the angle bisector of $\angle BPC$ and the perp bisector of $BC$ intersect in $A$ but it's well-known the intersection is on the circle $\odot BPC$ but $A,P$ are on the same side of $BC$ and $\angle BPC\not =\angle A$ a contradiction hence the conclusion.

Thanks, but can there be a proof with transformations?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Tumon2001

449 posts

Tumon2001

#6 h Feb 18, 2017, 1:59 AM • 2 Y

Y by arandomperson123, Adventure10

Another solution:

Let $MN$ be parallel to $BC$ . Then the result follows immediately as $AP$ is the perpendicular bisector of $BC$ .

Let $\angle APM$ = $\angle APN$ . Let $\omega$ be the circumcircle of $\Delta ABC$ .
Let $AP\cap \omega$ = $D$ .
Then, $\angle ADB$ = $\angle ADC$ .
Also, $\angle BPD$ = $\angle CPD$ .
So, triangles $BPD$ and $CPD$ are congruent.
Thus, $\Delta BPC$ is isosceles and so $AP$ is the perpendicular bisector of $BC$ .
Trivial angle chasing proves that $BMNC$ is cyclic. Hence, the result follows.

This post has been edited 1 time. Last edited by Tumon2001, Feb 18, 2017, 2:00 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

arandomperson123

430 posts

arandomperson123

#7 h Feb 18, 2017, 3:12 AM • 2 Y

Y by Adventure10, Mango247

Tumon2001 wrote:

Another solution:

Let $MN$ be parallel to $BC$ . Then the result follows immediately as $AP$ is the perpendicular bisector of $BC$ .

Let $\angle APM$ = $\angle APN$ . Let $\omega$ be the circumcircle of $\Delta ABC$ .
Let $AP\cap \omega$ = $D$ .
Then, $\angle ADB$ = $\angle ADC$ .
Also, $\angle BPD$ = $\angle CPD$ .
So, triangles $BPD$ and $CPD$ are congruent.
Thus, $\Delta BPC$ is isosceles and so $AP$ is the perpendicular bisector of $BC$ .
Trivial angle chasing proves that $BMNC$ is cyclic. Hence, the result follows.

Very nice solution, thank you very much!

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

rmtf1111

698 posts

rmtf1111

#8 h Feb 18, 2017, 8:19 AM • 1 Y

Y by Adventure10

Overkill for only if part