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Prove using vector methods that if a median in a triangle is perpendicular to the corresponding base, then the triangle is isosceles.

I know the basic idea. The dot product of the median vector and that of the base vector is $0$. Using that we have to prove the lengths of the sides containing the median are equal, but I don't know how to go about doing that.

Hints please.

Edit

My working: (as Blue suggested)

The equation for$\vec c$ is $ \vec r = (1-t)\vec a + t \vec b $ and the equation for $\vec m$ is $\vec r = \lambda\Big(\dfrac {\vec a + \vec b} 2\Big)$

Setting the dot product to zero gives this:

$$ \lambda/ 2 (1 + \vec a \cdot \vec b) = 0$$

What can I conclude from this?

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    $\begingroup$ Reflect the triangle in the median. $\endgroup$ – Daniel Fischer Jul 26 '13 at 18:07
  • $\begingroup$ I don't understand what you mean by 'Reflect the triangle'. $\endgroup$ – Parth Thakkar Jul 26 '13 at 18:08
  • $\begingroup$ @ParthThakkar He's saying that if it's the median, then the two opposing vertices must be the same distance from the median. The isometry that is the reflection on the median, as it's perpendicular to other base must swap those two vertices and leave the third invariant (it's contained in the median). Therefore the two other sides are the same length because the isometry sends one to the other. $\endgroup$ – MyUserIsThis Jul 26 '13 at 18:33
  • $\begingroup$ In your edit, writing "$\vec{r}=$" in when talking about both $\vec{c}$ and $\vec{m}$ is a little confusing, but that's beside the point. The expression $(1-t)\vec{a}+t\vec{b}$ is not $\vec{c}$; that's the parametric form of the complete line through the vector's endpoints. (Each value of $t$ gives a different point on that line.) Likewise, $\lambda \left(\vec{a}+\vec{b}\right)/2$, allowing $\lambda$ to vary, gives all the points on the line determined by the midpoint and the origin. I'll edit my answer to provide some clarity. $\endgroup$ – Blue Jul 26 '13 at 18:35
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You should be able to prove that the two triangles into which the median cuts the original triangle have one common side (the median), and they have another equal side - the base. The angle between the median and the base is equal in the two triangles.

Therefore we have equality of two sides and the included angle, which is sufficient to show that the triangles are congruent and the other corresponding sides and angles are equal.

To use vector methods, choose origin the midpoint of the base. Call the vector to the apex $\vec a$ and the vector to the right-hand base vertex $\vec b$. We then have $\vec a\cdot \vec b=0$.

The length of the right-hand side is $(\vec a - \vec b)\cdot(\vec a - \vec b)=\vec a \cdot \vec a - 2\vec a\cdot \vec b +\vec b \cdot \vec b=|a|^2+|b|^2$

The the vector to the left-hand base vertex is $-\vec b$ so the length of the left-hand side is similarly $(\vec a + \vec b)\cdot(\vec a + \vec b)$

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  • $\begingroup$ That's a nice way, but the thing is, this question requires us to use vectors. These were given as exercises in vectors - so we can't use the 'regular' geometry - moreover, I'd like to know how to do it the vector way. $\endgroup$ – Parth Thakkar Jul 26 '13 at 18:35
  • $\begingroup$ @ParthThakkar I added a different way of approaching things, which does it "the vector way". $\endgroup$ – Mark Bennet Jul 26 '13 at 18:43
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As a hint, think about where the median line intersects the base, using that information (and that which you already stated) you should be able to reach the conclusion.

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Edit. Re-written to review fundamental facts for OP's benefit.


Facts

  • The vector from point $P$ to point $Q$ is given by $\vec{PQ} = Q-P$.
  • The midpoint of $PQ$ is $\frac{1}{2}\left(P+Q\right)$

(To derive the second fact, note that one gets to the midpoint by starting at $P$ and traveling along the half-vector towards $Q$. This puts the midpoint at $$P + \frac{1}{2}\vec{PQ} = P + \frac{1}{2}\left(Q-P\right) = \frac{1}{2}\left( 2 P + Q - P \right) = \frac{1}{2}\left( P + Q \right)$$ as claimed.)


Now, let the $\triangle AOB$ have "base" $AB$, with $M$ the midpoint of that base. Situate vertex $O$ at the origin. From the above, we see that sides of the triangle, and the median in question, determine these vectors:

$$\vec{OA} = A - O = A \qquad \vec{OB} = B - O = B \qquad \vec{AB} = B - A$$ $$\vec{OM} = M-O = M = \frac{1}{2}\left( A + B \right)$$

We're investigating what happens when the median and the base are perpendicular, which corresponds to the condition:

$$\vec{OM}\cdot\vec{AB} = 0$$

Substituting our formulas for the vectors, this becomes

$$\frac{1}{2} \left( A + B \right)\cdot \left( B - A \right) = 0$$

Clearing the multiplied $\frac{1}{2}$ from both sides, we can expand the dot product and simplify thusly:

$$\begin{align} 0 &= A\cdot B - A\cdot A + B \cdot B - B\cdot A \\ &= |OB|^2 - |OA|^2 \end{align}$$

Consequently,

$$|OA|^2 = |OB|^2$$

and the triangle is isosceles.

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  • $\begingroup$ That's what I did, but I am not getting anything useful. Please wait, I'll just edit the question to show my working. $\endgroup$ – Parth Thakkar Jul 26 '13 at 18:21

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