Question

5: use a two - column or paragraph proof to prove that the diagonals of a rhombus meet perpendicularly.
given: abcd is a rhombus.
prove: ∠bmc (or any angle at m) is a right angle

Explanation:

Step1: Properties of a rhombus

In rhombus \(ABCD\), \(AB = BC = CD=DA\) (Definition of a rhombus). Also, the diagonals of a parallelogram bisect each other. Since a rhombus is a parallelogram, \(AM = CM\) and \(BM=DM\) (Diagonals of a parallelogram bisect each other).

Step2: Congruent triangles

Consider \(\triangle ABM\) and \(\triangle CBM\). \(AB = CB\) (Sides of a rhombus), \(AM = CM\) (Diagonals of a parallelogram bisect each other), \(BM = BM\) (Common side). So, \(\triangle ABM\cong\triangle CBM\) (SSS - Side - Side - Side congruence criterion).

Step3: Angles in congruent triangles

Since \(\triangle ABM\cong\triangle CBM\), \(\angle AMB=\angle CMB\).

Step4: Linear pair

\(\angle AMB+\angle CMB = 180^{\circ}\) (Linear pair of angles). Let \(\angle AMB = x\) and \(\angle CMB=x\) (from step 3). Then \(x + x=180^{\circ}\), \(2x = 180^{\circ}\), \(x = 90^{\circ}\).

Answer:

Statement	Justification
\(AB = BC\), \(AM = CM\), \(BM = BM\)	\(AB = BC\) (sides of a rhombus), \(AM = CM\) (diagonals of a parallelogram bisect each other), \(BM\) is common
\(\triangle ABM\cong\triangle CBM\)	SSS congruence
\(\angle AMB=\angle CMB\)	Corresponding parts of congruent triangles
\(\angle AMB+\angle CMB = 180^{\circ}\)	Linear pair
\(2\angle CMB=180^{\circ}\)	Substitution (\(\angle AMB=\angle CMB\))
\(\angle CMB = 90^{\circ}\)	Solving \(2\angle CMB=180^{\circ}\)