Question

recognizing sss situations
\overline{ab} \cong \overline{bc} and \overline{ad} \cong \overline{cd}
what additional information would make it immediately possible to prove that triangles axb and cxb are congruent using the hl theorem?
what additional information would make it immediately possible to prove that triangles axd and cxd are congruent using the sss congruence theorem?

Explanation:

Step1: Analyze triangles AXB and CXB

We know \( AB \cong BC \) (given) and \( XB \) is common to both triangles \( AXB \) and \( CXB \). For HL (Hypotenuse - Leg) congruence, we need a right angle. So if \( \angle AXB \) and \( \angle CXB \) are right angles (i.e., \( AC \perp BD \)), then by HL (since \( AB = BC \) (hypotenuse) and \( XB \) is common leg), triangles \( AXB \cong CXB \).

Step2: Analyze triangles AXD and CXD

We know \( AD \cong CD \) (given) and \( XD \) is common to both triangles \( AXD \) and \( CXD \). For SSS congruence, we need \( AX \cong CX \). If \( X \) is the midpoint of \( AC \) (i.e., \( AX = CX \)), then with \( AD = CD \) and \( XD = XD \), by SSS, triangles \( AXD \cong CXD \).

Answer:

For triangles \( AXB \) and \( CXB \) to be congruent by HL: \( AC \perp BD \) (i.e., \( \angle AXB=\angle CXB = 90^\circ \))
For triangles \( AXD \) and \( CXD \) to be congruent by SSS: \( AX = CX \) (i.e., \( X \) is the midpoint of \( AC \))