Show that AN=1 cm using the knowledge of equiangular triangles.

  • Context: MHB 
  • Thread starter Thread starter mathlearn
  • Start date Start date
  • Tags Tags
    Knowledge Triangles
Click For Summary

Discussion Overview

The discussion revolves around demonstrating that segment AN measures 1 cm using properties of equiangular triangles. Participants explore various geometric relationships and angle measures within the context of a specific problem involving triangles and their congruences.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks clarification on how triangles AMN and BMC can be considered equiangular, noting that they have one pair of equal angles.
  • Another participant hints at the relationship between angles AMN and BMC, suggesting they sum to 90 degrees.
  • Further contributions reiterate the angle relationships, with one participant asserting that angle MAN measures 90 degrees.
  • Some participants discuss the implications of angle measures and congruences, proposing that triangles AMN and BCM are similar based on their angle relationships.
  • Another participant states that both triangles are equiangular due to the presence of a common 90-degree angle and another pair of equal angles.
  • There are requests for inline image attachments to aid in visualizing the problem, emphasizing the importance of accessibility to the figures involved.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the equiangular nature of the triangles, with some agreeing on certain angle measures while others remain uncertain about the implications of these relationships. The discussion does not reach a consensus on the overall approach to proving that AN equals 1 cm.

Contextual Notes

Participants reference specific angle relationships and congruences without fully resolving the mathematical steps necessary to establish the final conclusion. There are indications of missing assumptions or definitions that could clarify the reasoning.

mathlearn
Messages
331
Reaction score
0
I need help with the fourth part of this series of the questions,

k15j47.jpg


Here's the copied figure

View attachment 5968

ii.
AM=MB (M midpoint)
NAM=MBP($90^\circ$)
AMN=BMP(vertically opposite)

$\therefore \triangle AMN \cong \triangle BMP \left(AAS\right)$iii.MN=MP (triangle AMN & triangle BMP congruent)
MC=MC (Common side)
NMC=CMP(Data)

$\triangle CMN \cong \triangle MCP \left(SAS\right)$

$\therefore \angle BCM= \angle MCN $

________________________________________________________________________

Now how are the triangles AMN & BMC equiangular ? One pair of equal angles would be

$\angle NAM$= $\angle MBC (90^\circ)$

Now what are the other two angles & what are the reasons for their equality ? (Thinking) (Smile)
 

Attachments

  • mathlearn.jpg
    mathlearn.jpg
    30 KB · Views: 146
Last edited:
Mathematics news on Phys.org
Hint: $\angle{AMN}+\angle{BMC}=90^\circ$
 
greg1313 said:
Hint: $\angle{AMN}+\angle{BMC}=90^\circ$

(Yes) Thanks,

$\angle{BMC}=90^\circ - \angle{AMN}$

I am not still seeing why they are equiangular. (Sweating)
 
mathlearn said:
I need help with the fourth part of this series of the questions...

Please upload images here as inline attachments rather than to some image hosting site. Those image hosting sites don't work very well, and some of us (like me) simply cannot see them.

When I come across these "invisible" images, then I have to follow the link to the image, download the image to my hard drive, resize/crop/rotate the image in many cases, save the image and then upload the image as an inline attachment so that everyone can see it.

While it is your prerogative to use spoiler tags, I don't see why you would hide an image that's part of the problem statement. Typically the spoiler tag is used to hide something that might give away a hint or solution that others may not wish to see before attempting a problem. :)
 
MarkFL said:
Please upload images here as inline attachments rather than to some image hosting site. Those image hosting sites don't work very well, and some of us (like me) simply cannot see them.

When I come across these "invisible" images, then I have to follow the link to the image, download the image to my hard drive, resize/crop/rotate the image in many cases, save the image and then upload the image as an inline attachment so that everyone can see it.

While it is your prerogative to use spoiler tags, I don't see why you would hide an image that's part of the problem statement. Typically the spoiler tag is used to hide something that might give away a hint or solution that others may not wish to see before attempting a problem. :)

My apologies for the inconvenience caused (Doh). I will keep that in mind (Smile), Now I guess it's time to get back to the math (Smile)
 
Well, part of being a staff member is doing things like this, and I really only meant my request to be considered for future posting only...and certainly not as an accusation of deliberately causing an inconvenience.

Another advantage of uploading attachments rather than linking to other sites is that those other sites may disappear or otherwise remove such images periodically as part of their routine housecleaning, and then that would leave us with threads that no longer have their accompanying images.

And yes...back to the important stuff...the math. :D
 
What is the measure of $\angle{MAN}$?
 
greg1313 said:
What is the measure of $\angle{MAN}$?

measure of $\angle{MAN}=90^\circ$ (Smile)
 
Some remarks on (iv) and (v):

(iv)

$$\angle{AMN}+\angle{CMB}=90^\circ$$

$$\angle{AMN}+\angle{ANM}=90^\circ\quad\text{(angles of a triangle sum to }180^\circ\text{)}$$

$$\implies\angle{CMB}=\angle{ANM}\text{ hence }\triangle{AMN}\sim\triangle{BCM}$$

(v)

$$\overline{AN}=\overline{BP}=1$$

$$\overline{CN}=\overline{CP}=5\quad\text{(}\triangle{CNP}\text{ is isosceles)}$$
 
  • #10
greg1313 said:
Some remarks on (iv) and (v):

(iv)

$$\angle{AMN}+\angle{CMB}=90^\circ$$

$$\angle{AMN}+\angle{ANM}=90^\circ\quad\text{(angles of a triangle sum to }180^\circ\text{)}$$

$$\implies\angle{CMB}=\angle{ANM}\text{ hence }\triangle{AMN}\sim\triangle{BCM}$$

(v)

$$\overline{AN}=\overline{BP}=1$$

$$\overline{CN}=\overline{CP}=5\quad\text{(}\triangle{CNP}\text{ is isosceles)}$$

Now both the triangles are equiangular! (Happy) Because one angle is 90 in both , & the other pair of equiangular angles are CMB & ANM.

(Smile) Thank you very much , If there's any name for the method that you have used in iv, May I please have the name? (Wasntme)