If I have a diagram, like the following:


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)

  • $\begingroup$ 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

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 $AB=BC$.


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