0
$\begingroup$

I found orthocentric triangle proof really confusing at angles part. There isn't really much online info about proving such a triangle, so there is no valid explanation of the process

This particular part is confusing to me:

enter image description here

The heights $AD$ and $BE$ are drawn in the acute-angled triangle ${\triangle}ABC$. Here, triangle ${\triangle}DCE$ is similar to triangle ${\triangle}ABC$. Because of the similarity, their angles have to be equal. But they also have to correspond each other, which is not happening in the proof:

enter image description here

Rotating triangle ${\triangle}ABC$, we see that in reality it's supposed to be ${\angle}DCE = {\angle}ACB$. But the proof states opposite: ${\angle}DEC = {\angle}ACB$ and ${\angle}CDE = {\angle}BAC$. Eventually we get this picture:

enter image description here

My questions is: How, if angles clearly do not correspond? Is there a mistake or I overlooked something?

$\endgroup$
8
  • $\begingroup$ You must show the images in your post. Right now readers have to click on the links to see each image. $\endgroup$ Commented Aug 12 at 13:16
  • $\begingroup$ This triangle is addressed in virtually every geometry book of the last 200 years. Flip the orientation of the little triangle on C and you will see the similarity makes sense. $\endgroup$ Commented Aug 12 at 13:16
  • $\begingroup$ @DuongNgo how if system does not permit me to do so? Have to paste a link $\endgroup$ Commented Aug 12 at 13:19
  • $\begingroup$ @RobinSparrow Did so and even showed on picture. The angles do not correspond as it seems $\endgroup$ Commented Aug 12 at 13:21
  • $\begingroup$ DCE is similar to ACB $\endgroup$ Commented Aug 12 at 13:27

1 Answer 1

Reset to default
1
$\begingroup$

First of all, that's not a proof, it's rather an image. Second, you just rotated $\Delta$ DEC in the incorrect way. You already proved that $\Delta ABC \sim \Delta DEC$, but that's because $90-\angle BAD=\angle ABC= \angle DEC$ and $\angle DCE=\angle BCA$ so for the first theorem of similitudine you'll get that $\Delta ABC \sim \Delta DCE$, and from that you'll have that AC : DC = DE : BA= EC : CB but in your rotation you are saying that AC: BA = DE : DC = EC : CB, you had rearrenged the sides in the wrong way, in the image you just to flip one of the figures on itself.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.