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The problem is as follows:

In the figure it is known $VN=NA$ and $AH+HM=MV$ and $VA=10\,cm$. Using this information find $AH$.

Sketch of the problem

The choices given in my book are:

$\begin{array}{cc} 1.&\textrm{6 cm}\\ 2.&\textrm{7 cm}\\ 3.&\textrm{8 cm}\\ 4.&\textrm{5 cm}\\ 5.&\textrm{4 cm}\\ \end{array}$

According to the answers sheet the answer is choice 4.

I really don't know how to solve this problem. Is there a way to solve it using congruence or some sort of construction?.

A while ago I posted a similar problem to this and I attempted to use that strategy but it didn't work for that problem.

Thus how to solve this using only euclidean geometry constructions?. Comparing with known triangles it can be allowed.

I tried looking for which lines to draw but I couldn't find the right one. Therefore can someone help me here please?.

Chris Steinbeck Bell
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1 Answers1

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enter image description here Let's draw a perpendicular line from the point $N$ to the line segment $VH$ and call its intersection with $VH$ $P$. Since $VN=NA$, $NP$ is the middle base. Accordingly, let $2NP=AH=2m$ and $HM=n$. So $VP=PH=m+n$ since $VM=2m+n$ and $NP$ is the middle base. The problem is now solved. $$PM=(m+n)-n=m,$$ $$NP=PM \Longrightarrow 3\alpha=45^{\circ},$$ $$3\alpha=45^{\circ} \Longrightarrow 2\alpha=30^{\circ},$$ $$2\times 2m =VA=10 \ (\text{According to the 30-60-90 triangle}) \Longrightarrow 2m=AH=5.$$

adzetto
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  • I recommend adding in your answer that $PM=m$ because you are using the information given as, $AH+HM=MV$. I called $PM=g$, then $AH=2m$, $HM=n$ and $MV=PM+PV$, as $MV=g+(g+n)$, then $2m+n=2g+n$, therefore $g=m$ so $PM=m$, and this solves the problem as $PM=PN$ and as it is a right triangle it follows your instructions. This will make your answer more easy to understand. – Chris Steinbeck Bell Jun 23 '22 at 08:26
  • By the way, can you help me with this question is it possible to solve that without using triangle similarity?. Maybe does it exist some construction?. Overall thanks in advance. – Chris Steinbeck Bell Jun 23 '22 at 08:30
  • Can you also help me here I am still stuck on that problem. I will appreciate your help. – Chris Steinbeck Bell Jun 23 '22 at 08:39