Right Triangle Geometry

For students of class 9-10 (age 14-16)
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Tasnood
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Right Triangle Geometry

Unread post by Tasnood » Fri Mar 02, 2018 7:49 pm

A famous and easy geometry problem:
Let triangle $\triangle{ABC}$ have a right angle at $C$, and let $M$ be the midpoint of the hypotenuse $AB$. Choose a point $D$ on line $BC$ so that angle $\angle{CDM}$ measures $30$ degrees. Prove that the segments $AC$ and $MD$ have equal lengths.

[Sorry if it was posted in the forum already]

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samiul_samin
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Re: Right Triangle Geometry

Unread post by samiul_samin » Sat Mar 03, 2018 1:17 am

Can I use this type of diagram to solve the problem?
Screenshot_2018-03-03-00-51-29-1.png

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Tasnood
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Re: Right Triangle Geometry

Unread post by Tasnood » Sat Mar 03, 2018 8:44 am

Perfect rational diagram it is :)

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samiul_samin
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Re: Right Triangle Geometry

Unread post by samiul_samin » Sat Mar 03, 2018 1:16 pm

Solution.
At first draw $ME\perp BC$.Then we get the following diagram:
Screenshot_2018-03-03-13-04-04-1.png

Now,$ME\parallel AC$,

So,$\dfrac {BM}{BA}=\dfrac {BE}{BC}=\dfrac {ME}{AC}=\dfrac 12$

$\Rightarrow AC=2ME$... ... ... ($1$)

And,in the right triangle $\triangle DEM$

$\dfrac {ME}{MD}=sin 30^\circ=\dfrac 12$

$\Rightarrow MD=2ME$... ... ...($2$)

From ($1$) & ($2$), $AC=MD$ [Q.E.D]

aritra barua
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Re: Right Triangle Geometry

Unread post by aritra barua » Sat Mar 03, 2018 4:37 pm

Let $F$ be such a point so that $ACDF$ is a rectangle.So,angle $MDF$=$60$°.Since $M$ is the circumcenter of $\bigtriangleup ABC$,$MC=MA$.Now,using perpendicularity lemma $MC^2+MF^2$=$MA^2+MD^2$ implying $MD=MF$.Since,angle $MDF$=$60$°,$\bigtriangleup MDF$ is equilateral.

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Tasnood
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Re: Right Triangle Geometry

Unread post by Tasnood » Sat Mar 03, 2018 5:44 pm

So many solutions :!:

Applying Sine law in $\triangle CDM$, we get:
$\frac {CM}{sin \angle CDM}=\frac{DM}{sin \angle DCM} \Rightarrow \frac{CM}{\frac {1}{2}}=\frac{DM}{sin \angle DCM} \Rightarrow 2CM=\frac{DM}{sin \angle DCM} \Rightarrow DM=2CM \times sin\angle DCM$ ...(1)

$M$ is the midpoint of the hypotenuse $AB$. So, $M$ is the center of the circumcircle of $\triangle ABC$.
$AM=BM=CM=\frac{1}{2}AB \Rightarrow 2CM=AB$

In $\triangle BCM$, $BM=CM$. So $\triangle BCM$ is isosceles.
In $\triangle BCM$, $\angle MBC=\angle MCB= \angle MCD$

Putting the values properly, we get:
$DM=2CM \times sin\angle DCM \Rightarrow DM=AB \times sin \angle MBC \Rightarrow DM=AB \times {\frac {AC}{AB}} \Rightarrow AC$ [According to $\triangle BCM$] :D

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samiul_samin
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Re: Right Triangle Geometry

Unread post by samiul_samin » Sat Mar 03, 2018 7:32 pm

I am thinking to solve it by using co-ordinate.But I am not getting any that type solution.Can anyone help me?

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samiul_samin
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Re: Right Triangle Geometry

Unread post by samiul_samin » Mon Feb 18, 2019 11:35 pm

Tasnood wrote:
Sat Mar 03, 2018 8:44 am
Perfect rational diagram it is :)
Screenshot_2019-02-18-23-33-08-1.png
Screenshot_2019-02-18-23-33-08-1.png (10.7 KiB) Viewed 430 times

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