0

The problem is as follows:

In an acute angled triangle $ABC$ $(AB<BC)$, the perpendicular bisector mediatrix of $AC$ intersects $BC$ at point $N$, and the height $BH$ intersects $AN$ at point $E$, If $\textrm{AE = 3 m}$ and $\textrm{BC = 13 m}$. Find $BN$.

The alternatives in my book are:

$\begin{array}{ll} 1.&\textrm{5 m}\\ 2.&\textrm{4 m}\\ 3.&\textrm{4.5 m}\\ 4.&\textrm{3.5 m}\\ \end{array}$

This problem did not come with a diagram or a drawing so the best that I could do is to make an interpretation of what it is being mentioned and this is seen below:

Sketch of the problem

Now here it comes the part on How do I use properly the information which has been mentioned?

First off is my problem with this word: mediatrix, it seems that the aluded word seems to indicate that it is a median and an angle bisector at the same time. But is this okay?. Or is it just a perpendicular bisector and only that?.

Other than that,

I have currently ran out of ideas.

The only thing which had come to my mind is that when a small angle opposes to one particular side that side must be also small when compared to a greater angle. But how does this can be used to solve this problem?.

Since I'm not good on this subject. It would help me a lot if someone could help me here?.

The intended approach I am looking is to solve this relying only in euclidean geometry.

Is my drawing a correct interpretation?. If not please feel free to correct it in such a way that it is right.

Please if you can try to include a drawing in your answer because that could help me better to spot if a construction is needed.

Chris Steinbeck Bell
  • 4,175

1 Answers1

1
  1. It is trivial to show $AN=NC$
  2. Slightly more complicated, but still easy $\angle NBE=\angle NEB$, therefore $NE=NB$.
  3. In your figure, move the "3m" label onto $AE$, not $EN$ Then $$BN+NC=13\\AN-NE=AE=3\\NC-BN=3$$ Therefore $BN=5$

EDIT Here are some additional information that was pointed out in the comments: In step 2, I used $\angle NCA=\angle CAN$ from step 1, then $\angle AEH=90^\circ-\angle CAN$. Opposite angles $\angle AEH$ and $\angle BEN$ are equal. In the right angle triangle $BCH$, the angle $\angle CBH=90^\circ-\angle BCH$. Therefore $\angle CBH=\angle BEN$, so $\triangle BEN$ is isosceles. You can get to the same conclusion if you draw a parallel to $AC$ through $N$, that intersects $BH$ at $M$. You know $MN\perp BH$, and then you look to prove $\angle MNE=\angle MNB$, to show that the two triangles are congruent and $EN=NB$.

The other question that showed up was why is it necessary for the triangle to be acute? It does not need to be (with caveats). Angle $\angle ABC$ can be obtuse. To prove this, start with a very obtuse isosceles triangle $ANC$, where $AN=NC=8$. Now choose point $E$ on $AN$ such that $AE=3$ and $EN=5$. Draw the perpendicular from $E$ to $AC$. Point $B$ is at the intersection of $NC$ and $EH$. You can arbitrarily increase angle $\angle ANC$ towards $180^\circ$, and then $\angle ABC$ will tend to the same value. However, if angle $BAC$is obtuse, the geometry will change. Angle $BCA$ can't be obtuse and have $AB<BC$

Andrei
  • 37,370
  • Howdy. I've tried to explore different methods to find how $\angle NBE=\angle NEB$ but I cannot do it on my own. Can you help me on how did you concluded this Did you used congruence or similarity?. The rest I can understand, but it is the second statement in your answer which I'm stuck. Therefore, can you include an explanation on how you concluded that those angles are the same?. – Chris Steinbeck Bell Jan 15 '21 at 01:36
  • Draw a parallel to $AC$ through $N$, and call $M$ the intersection with $BH$. Then look at the pair of angles that are equal when you intersect these parallel lines with $AN$ and with $BC$. Alternatively, $BEN=AEH$. Then $AEH=90-EAH=90-ACN=90-(90-NBE)=NBE$. Both ways, you get $NEB$ is isosceles. – Andrei Jan 15 '21 at 01:46
  • I tried to follow your directions, but I still need assistance. I drew the parallel through $N$, but what pair of angles are you referring? I could spot that $\angle MNE=\angle EAH$ and $\angle MNB = \angle ACN$ both are the same, and as a the perpendicular makes an angle bisector, therefore $\triangle BEN$ is isosceles. Where you referring to this?. – Chris Steinbeck Bell Jan 15 '21 at 09:01
  • Your second method it is confusing. Yes $\angle BEN=\angle AEH$. But I guess what you meant to say is, EAH=90-AEH and ACN=EAH and because of this reason $HBC=90-(90-AEH)$, hence $HBC=AEH$. Were you referring to this?. Can you confirm?. – Chris Steinbeck Bell Jan 15 '21 at 09:02
  • In both cases you want to show that $\triangle NBE$ is isosceles, such that $NB=NE$. So your first comment was exactly what I had in mind. For the second comment, you just missed the last step: $AEH=BEN$. So $HBC=BEN$, therefore $BEN$ is isosceles. – Andrei Jan 15 '21 at 15:04
  • Sorry, for brevity purposes as space is limited in comments I did not included the last step which you mentioned but I arrived to the same conclusion as you. Maybe you want to move these comments to be part of you answer?. Because I feel these are crucial to understand the reason of why is it an isosceles, as this would solve the problem. Again, this does answer the problem, but the initial question remains unattended, what is necessary to use the fact that this triangle was acute angled?. From your method, it seems it was irrelevant, but I disagree. – Chris Steinbeck Bell Jan 16 '21 at 01:23
  • @ChrisSteinbeckBell I've modified the solution to answer your comments. Angle $ABC$ can be obtuse, and still get the same solution. – Andrei Jan 16 '21 at 07:22
  • @ChrisSteinbeckBell Did the extra edits answered your questions? If yes, don't forget to accept the answer – Andrei Jan 18 '21 at 22:54
  • Thanks for adding that. Sure! I've accepted your answer. – Chris Steinbeck Bell Jan 20 '21 at 10:36