2
$\begingroup$

If I have a diagram, like the following:

Diagram

And I want to make make a proof for something like how segment AB is $\cong$ to segment AC if segment BD $\cong$ segment DC (using Perpendicular Bisector Theorem), well to do this I would need to show that segment AD $\bot$ segment DC (or segment BD).

Can I state that it is a given that the two are $\bot$ because a right angle is shown (so this would be given), or do I need to say, first, that m$\angle$ADC = right $\angle$ (given), and then say that segment AD $\bot$ segment DC (def of $\bot$ lines)

$\endgroup$
  • $\begingroup$ Well, what, exactly, is known here? If all you are given is that "BC≅ DC" then it isn't true. In order to conclude that "AC≅ AC" you have to be [b]given[/b] that AD is perpendicular to BC. $\endgroup$ – user247327 Jun 2 at 0:40
  • $\begingroup$ @user247327 I am saying that it is given that $\angle$ADC is a right angle, by the way, might want to check your point names in your comment :] $\endgroup$ – BeastCoder2 Jun 2 at 0:44
  • $\begingroup$ Presumably you mean $BD=DC$ rather than $BC=DC$? $\endgroup$ – Henning Makholm Jun 2 at 0:48
  • $\begingroup$ @HenningMakholm Wow! I can't believe I didn't catch that, haha! Well, I fixed it now! $\endgroup$ – BeastCoder2 Jun 2 at 0:50
  • $\begingroup$ If the right angle (the yellow square) is given, then you can say that $BC\perp AD$. $\endgroup$ – Julian Mejia Jun 2 at 1:01
1
$\begingroup$

So, If the right angle (the rectangle in yellow) is given. I think is okay to say that $AD\perp BC$. Now, you say that then $AB=AC$ iff $BD=BC$ by using the perpendicular bisector theorem.

Even though this is fine, I think this is overkilling since it feels that the thing you want to prove follows trivially from this theorem.

Another way to prove this is by using Pythagorean theorem. $BD^2+AD^2=AB^2$ and $ DC^2+AD^2=AC^2$. If you substract these two equalities, you get $BD^2-DC^2=AB^2-AC^2$. So, $BD=DC$ iff $AB=BC$.

$\endgroup$

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.