Question

3.11 rhombuses (or rhombi, if you like to be greek)

1. rhombuses have a lot of properties, but a lot of them come from the fact that a rhombus is a special kind of parallelogram. the new properties are:

we need to start with some congruent triangles to prove the properties.

1. given: abdc is a rhombus.

prove: $\triangle acf \cong \triangle dcf$

1. what other triangles are congruent to $\triangle acf$ and $\triangle dcf$

1. mark up the diagram to show everything that is congruent in this picture. there is a lot!

Explanation:

Part 2: Proving $\boldsymbol{\triangle ACF \cong \triangle DCF}$

Step 1: Recall Rhombus Properties

In a rhombus, all sides are congruent, and opposite sides are parallel. Also, the diagonals of a rhombus bisect the angles and are perpendicular bisectors of each other. For rhombus \(ABDC\), \(AC = DC\) (all sides of a rhombus are equal).

Step 2: Identify Common Side

\(CF\) is a common side to both \(\triangle ACF\) and \(\triangle DCF\), so \(CF = CF\) (reflexive property of congruence).

Step 3: Angle Bisector Property

In a rhombus, the diagonal bisects the angles. So, \(\angle ACF = \angle DCF\) (diagonal \(CD\)? Wait, no—diagonal \(AD\)? Wait, in rhombus \(ABDC\), sides \(AC = DC\), and diagonal \(CF\) (wait, the diagonal here is \(AD\)? Wait, the diagram: \(ABDC\) is a rhombus, so \(AC = DC\), \(CF\) is common, and \(\angle ACF = \angle DCF\) (since diagonal bisects the angle). Wait, actually, in rhombus \(ABDC\), \(AC = DC\) (sides of rhombus), \(CF = CF\) (common), and \(\angle ACF = \angle DCF\) (diagonal bisects the vertex angle). Thus, by SAS congruence:

• \(AC = DC\) (sides of rhombus)
• \(\angle ACF = \angle DCF\) (diagonal bisects the angle)
• \(CF = CF\) (reflexive property)

Filling the Proof Table:

Statements	Reasons
2. \(AC = DC\)	All sides of a rhombus are congruent
3. \(\angle ACF = \angle DCF\)	Diagonal of a rhombus bisects the vertex angle
4. \(CF = CF\)	Reflexive property of congruence
5. \(\triangle ACF \cong \triangle DCF\)	SAS (Side-Angle-Side) Congruence Postulate

Part 3: Other Congruent Triangles

Since \(\triangle ACF \cong \triangle DCF\), by symmetry and rhombus properties, \(\triangle ACF \cong \triangle DCF \cong \triangle ABF \cong \triangle DBF\) (because the other diagonal and sides will also satisfy congruence via SAS or SSS, as the rhombus diagonals bisect each other and all sides are equal).

Part 4: Marking Congruent Parts

• Mark all sides: \(AC = DC = AB = DB\) (all sides of rhombus).
• Mark angles: \(\angle ACF = \angle DCF = \angle ABF = \angle DBF\), \(\angle CAF = \angle CDF = \angle BAF = \angle BDF\).
• Mark diagonals: \(AF = DF\) (diagonals bisect each other in a rhombus), \(CF = BF\) (diagonals bisect each other).
• Mark triangles: \(\triangle ACF \cong \triangle DCF \cong \triangle ABF \cong \triangle DBF\), and also \(\triangle ACD \cong \triangle ABD\) (SSS, as all sides equal).

Answer:

(for Part 2 Proof):
The proof uses the SAS Congruence Postulate with \(AC = DC\), \(\angle ACF = \angle DCF\), and \(CF = CF\). The table is filled as above.

For Part 3, triangles congruent to \(\triangle ACF\) and \(\triangle DCF\) are \(\boldsymbol{\triangle ABF}\) and \(\boldsymbol{\triangle DBF}\) (and each other, so \(\triangle ACF \cong \triangle DCF \cong \triangle ABF \cong \triangle DBF\)).

For Part 4, mark all sides as equal, angles formed by diagonals as equal, and triangles as congruent (e.g., tick marks on sides, angle arcs, etc.).