Question

triangles abc and dbc have the following characteristics:
bc is a side of both triangles
∠acb and ∠dcb are right angles
ac ≅ dc
which congruence theorem can be used to prove △abc ≅ △dbc?
sss
sas
aas
hl

Explanation:

Step1: Analyze the given information

We have two right - angled triangles (since \(\angle ACB\) and \(\angle DCB\) are right angles). The hypotenuse \(BC\) is common to both triangles (\(BC = BC\) by reflexive property) and one of the legs \(AC\cong DC\) (given).

Step2: Recall the congruence theorems

• SSS (Side - Side - Side): Requires three pairs of equal sides. We only know about two sides (\(BC\) and \(AC/DC\)) for sure in the context of the given information related to right - angled triangles.
• SAS (Side - Angle - Side): For non - right - angled triangles, but we have right - angled triangles here and a more specific theorem for right - angled triangles.
• AAS (Angle - Angle - Side): We are not given two pairs of equal angles in a non - right - angled triangle context.
• HL (Hypotenuse - Leg): For right - angled triangles, if the hypotenuse and one leg of a right - angled triangle are equal to the hypotenuse and one leg of another right - angled triangle, then the two right - angled triangles are congruent. Here, \(BC\) (hypotenuse) is common (\(BC\cong BC\)) and \(AC\cong DC\) (leg).

Answer:

HL