1
$\begingroup$

Two triangles $\triangle ABC$, $\triangle A'BC$ have the same base and the same height. Through the point $P$ where their sides intersect we draw a straight line parallel to the base; this line meets the other side of $\triangle ABC$ in X and the other side of $\triangle A'BC$ in $X'$. Prove that $XP = X'P$.

$\endgroup$
1
$\begingroup$

Observe that $\triangle XPA \sim \triangle ABC$ and $\triangle X'PA' \sim \triangle A'BC$. As the $XX'$ line is parallel to $BC$, and $\triangle ABC$ and $\triangle A'BC$ were of the same height, then $\triangle XPA$ and $\triangle X'PA'$ are also of the same height. The result follows from the fact, that the bases have the same length, that is $|XP| = |X'P|$ because $|BC| = |BC|$.

I hope this helps ;-)

Edit: Solution using trapezium area (as asked for in the comments):

We start with the area for trapezium $$|ABCA'| = |BCP| + |A'AP| + |BXP| + |CX'P| + |XPA| + |X'PA'|$$

then relate the areas of two triangles of the same height \begin{align} |BCP| + |BXP| + |XPA| + |A'AP| &= |ABC| \\ = |A'BC| &= |BCP| + |CX'P| + |X'PA'| + |A'AP| \end{align}

and combining both we get $$|BXP| + |XPA| = |CX'P| + |X'PA'|.$$

However $\triangle BXP$ and $\triangle CX'P$ are of the same height as well as $\triangle XPA$ and $\triangle X'PA'$. Writing this down we obtain

$$\frac{1}{2}b_1*(h_1+h_2) = \frac{1}{2}b_2*(h_1+h_2)$$

and after simplification ($h_1+h_2 \neq 0$)

$$b_1 = b_2.$$

$\endgroup$
  • $\begingroup$ Hi, I am working through a book and similarity is introduced in the next chapter. So is it possible to do this without using similarity? $\endgroup$ – user84973 Jul 4 '13 at 9:09
  • $\begingroup$ Do you know this theorem (sometimes called Tales' theorem, sometimes intercept theorem)? If yes, then apply it two times to get the solution. In fact similarity and the aforementioned theorem are roughly about the same underlying laws. $\endgroup$ – dtldarek Jul 4 '13 at 9:16
  • $\begingroup$ Sorry, I don't know this theorem. I should have mentioned book gives hint "consider the area of trapezium". I have considered the area in 3 ways, but it doesn't seem to simply to the result. $\endgroup$ – user84973 Jul 4 '13 at 9:25
  • $\begingroup$ I always find it funny to consider area before similarity (I mean the books), nevertheless, see the edits. $\endgroup$ – dtldarek Jul 4 '13 at 9:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.