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Question: In $\triangle A B C, D$ is the midpoint of $B C$ and $E$ is the foot of the perpendicular from $B$ on $A C$. Find $D E$ in terms of the angles and sides of $\triangle A B C$.


My attempt

Let $DE=l, EC=m, \angle ACB= x, \angle BAC= y$, after trigonometry bashing I got: $A E=\tan (90-y) \times 2 l \sin x$, and $BE= 2l \sin x$. I'm getting a lot of results from bashing but not getting a correct path to actually solve the problem.. Edit: Please note the question asks to find value of $DE$ in terms of side length as well as angles of the triangle

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    $\begingroup$ $DE=\frac{BC}{2}$ $\endgroup$
    – Lion Heart
    Commented Jun 5, 2022 at 22:15
  • $\begingroup$ @LionHeart yeah, I figured it out but still, how should I proceed $\endgroup$ Commented Jun 5, 2022 at 22:33
  • $\begingroup$ @Euclid_Euler You misunderstand what "in terms of the..." means in a problem. It means "in terms of nothing else but the...". It does not mean "in terms of all of the...". $\endgroup$
    – dxiv
    Commented Jun 5, 2022 at 23:05

2 Answers 2

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Since triangle $BEC$ is right, we know that point $D$ is the midpoint of the hypotenuse of a right triangle. It is well known that the circumcenter of a right triangle is at the midpoint of the hypotenuse, which is point $D$, which means that the length of $DE$ must be equal to the lengths of $BD$ and $CD$. Thus, the length of $DE$ is half of $BC$.

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  • $\begingroup$ It said to find DE in terms of angles and side length of the triangle $\endgroup$ Commented Jun 5, 2022 at 22:36
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    $\begingroup$ @Euclid_Euler I don't see a problem? BC is a side of the triangle... You don't need the angles $\endgroup$
    – Fatso Boo
    Commented Jun 5, 2022 at 22:41
  • $\begingroup$ then why does the question mention in terms of angles though? Where's angles involved in here. I'm confused $\endgroup$ Commented Jun 5, 2022 at 22:43
  • $\begingroup$ $DE = \frac{BC}2+(180^\circ-\alpha-\beta-\gamma)$. $\endgroup$
    – Mateo
    Commented Jun 5, 2022 at 22:46
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    $\begingroup$ @Euclid_Euler just because it says in terms of angles and sides does not mean you need angles and sides in your answer. In this case, we only used 1 side. $\endgroup$
    – Fatso Boo
    Commented Jun 5, 2022 at 22:55
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@Lion Heart gave a proper answer. The triangle BEC is right and $D$ is a midpoint of a hypotenuse.

Hint

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  • $\begingroup$ I figured it out, although forgot to mention it in the post, still how would it help? And details on it? $\endgroup$ Commented Jun 5, 2022 at 22:34
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    $\begingroup$ You've got the answer and ask how would it help? $\endgroup$
    – Mateo
    Commented Jun 5, 2022 at 22:48
  • $\begingroup$ That's not the right answer. Please read the last line of the question carefully. It says in terms of angles as well as sides of the triangle. Your answer neither involves the 3 sides, nor the angles. Explain how is it correct in any way? $\endgroup$ Commented Jun 5, 2022 at 22:51
  • $\begingroup$ Hey, @Euclid_Euler, everybody tells you that this is a proper answer and they are right. If you don't believe, maybe this answer will be acceptable for you: $DE = \frac{BC}2+(AC-BA)\cdot 0 + (180^\circ-\alpha-\beta-\gamma)$. $\endgroup$
    – Mateo
    Commented Jun 5, 2022 at 23:38

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