# Is it given that two lines are perpendicular if a right angle is shown?

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)

• 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. – user247327 Jun 2 at 0:40
• @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 :] – BeastCoder2 Jun 2 at 0:44
• Presumably you mean $BD=DC$ rather than $BC=DC$? – Henning Makholm Jun 2 at 0:48
• @HenningMakholm Wow! I can't believe I didn't catch that, haha! Well, I fixed it now! – BeastCoder2 Jun 2 at 0:50
• If the right angle (the yellow square) is given, then you can say that $BC\perp AD$. – 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.
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 $$AB=BC$$.