Question

here is rhombus abcd. complete the proof that in rhombus abcd the diagonal bd bisects angle abc. choose your answers from the drop - down menus. a. a rhombus has 4 congruent sides, so segments ab, cb, ad, and cd are congruent. b. segment db because segments are congruent to themselves. 2 pairs of opposite sides congruent, 2 pairs of adjacent sides congruent c. triangle dab is 4 congruent sides side - side triangle congruence theorem.

Explanation:

Part a

Step1: Recall rhombus properties

A rhombus is defined as a quadrilateral with 4 congruent sides. So the correct property for a rhombus here is "4 congruent sides".

Step2: Confirm the relation

Since a rhombus has 4 congruent sides, segments \(AB\), \(CB\), \(AD\), and \(CD\) are congruent (as all sides of a rhombus are equal).

Part b

Step1: Analyze triangle congruence

In rhombus \(ABCD\), when we draw diagonal \(BD\), we consider triangles \(DAB\) and \(DCB\) (or related triangles). But for the sides adjacent to the diagonal, in a rhombus, 2 pairs of adjacent sides are congruent? Wait, no—actually, in a rhombus, all sides are congruent, but when looking at the diagonal \(BD\), triangles \(ABD\) and \(CBD\) have sides: \(AB = CB\), \(AD=CD\), and \(BD\) is common. But the key here is the property of the rhombus sides. Wait, the drop - down for part b: the options are "2 pairs of opposite sides congruent" or "2 pairs of adjacent sides congruent". In a rhombus, all sides are congruent, so both pairs of opposite sides are congruent and both pairs of adjacent sides are congruent. But in the context of the proof (showing diagonal bisects the angle), we use the fact that in rhombus \(ABCD\), \(AB = BC\) and \(AD=DC\) (adjacent sides? Wait, no, \(AB\) and \(AD\) are adjacent, \(AB\) and \(BC\) are adjacent. Wait, actually, in a rhombus, 2 pairs of adjacent sides are congruent (since all four sides are equal, so \(AB = BC\), \(BC = CD\), \(CD = DA\), \(DA = AB\) – so adjacent sides are congruent). Wait, but the other option is "2 pairs of opposite sides congruent" which is also true for a rhombus (since it's a parallelogram). But in the context of the proof, when we have diagonal \(BD\), and triangles \(ABD\) and \(CBD\), we have \(AB = CB\) (adjacent sides) and \(AD = CD\) (adjacent sides). So the correct option is "2 pairs of adjacent sides congruent" because we are looking at the sides adjacent to the vertices of the angle being bisected.

Part c

Step1: Recall triangle congruence theorems

We have triangles \(DAB\) and \(DCB\) (or \(DAB\) and \(DBC\)? Wait, in rhombus \(ABCD\), diagonal \(BD\) splits it into two triangles \(ABD\) and \(CBD\). We know \(AB = CB\), \(AD = CD\), and \(BD=BD\) (common side). So by the Side - Side - Side (SSS) congruence theorem, the triangles are congruent. So the triangle congruence theorem here is SSS (Side - Side - Side).

Final Answers (for each part):

a. 4 congruent sides
b. 2 pairs of adjacent sides congruent
c. Side - Side - Side (SSS) Triangle Congruence Theorem

Answer:

Step1: Recall triangle congruence theorems

We have triangles \(DAB\) and \(DCB\) (or \(DAB\) and \(DBC\)? Wait, in rhombus \(ABCD\), diagonal \(BD\) splits it into two triangles \(ABD\) and \(CBD\). We know \(AB = CB\), \(AD = CD\), and \(BD=BD\) (common side). So by the Side - Side - Side (SSS) congruence theorem, the triangles are congruent. So the triangle congruence theorem here is SSS (Side - Side - Side).

Final Answers (for each part):

a. 4 congruent sides
b. 2 pairs of adjacent sides congruent
c. Side - Side - Side (SSS) Triangle Congruence Theorem