0
$\begingroup$

Prove that there cannot exist a triangle in which the trisectors of an angle also trisect the opposite side by using proportions.

Attempt I have started the proof by way of contradiction.

Suppose we have a triangle $\Delta BAC$ with trisectors $AD$ and $A$E so that $D$ and $E$ are points lying in side $BC$. I also supposed that $D$ and $E$ trisect $BC$ such that $BD=DE=EC$. I then looked at triangle $BAE$ and observed that $AD$ bisects $\angle BAE$ and that point $D$ is the midpoint of $BE$ therefore $BD$ is both a median and an angle bisector.

From this I get stuck and do not know how to apply proportions. I know we proceed similarly when looking at $\Delta DAC$.

$\endgroup$
1
$\begingroup$

It is enough to show (through the sine theorem, for instance) that in the following configuration the red segments are longer than the blue ones: enter image description here

$\endgroup$
0
$\begingroup$

You have a $\triangle{ABC}$. Let us say that side $BC$ is trisected by points $D, E$ such as $BD=DE=EC$. Note that $A_{\triangle{ACE}}=A_{\triangle{AED}}=A_{\triangle{ADB}}$ (equal bases, same height). Using area of a triangle formula this leads to $AC=AD$, $AE=AB$. But then $AE$ is a median and height of $\triangle{ACD}$ and $AD$ is a median and height of $\triangle{AED}$. We have two perpediculars from one point $A$ to side $BC$ that go to two distinct points. This is impossible.

$\endgroup$

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.