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

[mathlearn]

I need help with the fourth part of this series of the questions,

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 $

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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)

 

[Greg]

Hint: $\angle{AMN}+\angle{BMC}=90^\circ$

 

[mathlearn]

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)

 

[MarkFL]

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. :)

 

[mathlearn]

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)

 

[MarkFL]

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

 

[Greg]

What is the measure of $\angle{MAN}$?

 

[mathlearn]

What is the measure of $\angle{MAN}$?

measure of $\angle{MAN}=90^\circ$ (Smile)

 

[Greg]

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)

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

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

 

[mathlearn]

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)