Help, H.S. Math - Bisector congruence.

[Alpha6]

Originally I chose Angle-Side-Angle, but I think it is d, all of the above.

ASA because BDA and CDA are both right angles, BAD and CAD are congruent angles, and AD = itself.

SAS because BD = CD and

SSS because if it is a right angle that would mean AB = AC due to Pythagorean theorem.


Here is why I'm second guessing myself... When I see Khan Academy perform this, Khan never states Angle D is a right angle. Also he never states line BD = line CD, which confuses me because I thought the bisector splits the triangle in half and creates a right angle.

Help please

 

[Dr.Peterson]

Originally I chose Angle-Side-Angle, but I think it is d, all of the above.

ASA because BDA and CDA are both right angles, BAD and CAD are congruent angles, and AD = itself.

SAS because BD = CD and

SSS because if it is a right angle that would mean AB = AC due to Pythagorean theorem.

Here is why I'm second guessing myself... When I see Khan Academy perform this, Khan never states Angle D is a right angle. Also he never states line BD = line CD, which confuses me because I thought the bisector splits the triangle in half and creates a right angle.

I agree with you. They have given more information than is needed to prove that the triangles are congruent, so that it can be done in several ways. The fact that AD is the perpendicular bisector implies the right angle and the equal segments, allowing for SAS, and the fact that it is the angle bisector implies the congruent angles you need for ASA. By the Pythagorean Theorem, you can relate all the sides; that might be considered too indirect to count, so (c) might be considered the correct answer, but (d) is sensible. Note that only II requires the angle bisector.

I imagine whatever you saw in Khan Academy, it was not this same problem, but one that gave less information.