Question

which congruence theorem can be used to prove $\triangle bda \cong \triangle bdc$?
options: hl, ssa, aas, sss

Explanation:

Step1: Analyze the given triangles

We have right - angled triangles \(\triangle BDC\) and \(\triangle BDA\) (assuming \(BD\) is perpendicular to \(AC\) as \(D\) is a point on \(AC\) and from the congruence markings).

Step2: Identify the corresponding sides

In right - angled triangles \(BD\) is the common leg (\(BD = BD\)) and \(BC=BA\) (given by the congruence markings).

Step3: Apply the HL (Hypotenuse - Leg) theorem

For right - angled triangles \(\triangle BDA\) and \(\triangle BDC\), the hypotenuse \(BA = BC\) and the leg \(BD=BD\). By the HL (Hypotenuse - Leg) congruence theorem for right - angled triangles, \(\triangle BDA\cong\triangle BDC\).

SSA is not a valid congruence theorem in general (except for right - angled triangles where it is equivalent to HL). AAS (Angle - Angle - Side) requires two angles and a non - included side. SSS (Side - Side - Side) requires all three sides of one triangle to be equal to the corresponding sides of the other triangle. Here, we have right - angled triangles with hypotenuse and a leg equal, so HL is the correct theorem.

Answer:

A. HL