Question

3 in the figure on the right, given the diagonals of parallelogram abcd intersect perpendicularly, prove that abcd is a rhombus as follows. fill the appropriate words in the blanks.
way of thinking
to prove that abcd is a rhombus, use the definition that the four sides are equal.
if ab = ad, ab = bc = cd = da due to the condition of a parallelogram that a pair of opposite sides are equal in length.
proof
let o be the intersection point of the diagonals.
in △aob and △aod,
given: ∠aob = ∠□ = 90° ……… ①
since the diagonals of a parallelogram intersect at the midpoint of each,
bo = □ ……… ②
also, ao is shared ……… ③
from ①, ②, and ③, since the two □ and the included □ are equal respectively,
△aob ≅ △□
therefore, ab = □
since the opposite sides of the parallelogram are equal, the four sides of parallelogram abcd are equal. therefore, parallelogram abcd is a rhombus.
4 in the figure on the right, given perpendiculars ae and af are drawn from vertex a of rhombus abcd to sides bc and cd respectively, prove be = df as follows. fill the appropriate words in the blanks.
way of thinking
prove that △abe and △adf are congruent.
proof
in △abe and △adf,
given: ∠aeb = ∠□ = 90° ……… ①
∠abe = ∠adf ……… ②
from the definition of a rhombus,
ab = □ ……… ③
from ①, ②, and ③, since the □ and one □ are equal respectively,
△□ ≅ △□
therefore, be = □

Explanation:

Problem 3 (Proving ABCD is a rhombus)

Step1: Identify the right angle

In parallelogram \(ABCD\), diagonals intersect at \(O\) and are perpendicular. So \(\angle AOB=\angle AOD = 90^\circ\) (since diagonals intersect perpendicularly, both angles at \(O\) between the diagonals are right angles).

Step2: Midpoint property of parallelogram diagonals

Diagonals of a parallelogram bisect each other, so \(BO = DO\) (because \(O\) is the midpoint of diagonal \(BD\)).

Step3: Congruence of triangles

We have \(AO\) common (shared side), \(BO = DO\) (from step 2), and \(\angle AOB=\angle AOD = 90^\circ\) (from step 1). By SAS (Side - Angle - Side) congruence criterion, \(\triangle AOB\cong\triangle AOD\) (two sides and included angle are equal: \(AO = AO\), \(BO = DO\), \(\angle AOB=\angle AOD\)).

Step4: Equal sides from congruence

Since \(\triangle AOB\cong\triangle AOD\), corresponding sides \(AB = AD\). In a parallelogram, opposite sides are equal (\(AB = CD\), \(AD = BC\)), so \(AB=BC = CD = DA\), hence \(ABCD\) is a rhombus.

Filling the blanks:

• First blank: \(AOD\)
• Second blank: \(DO\)
• Third blank: sides (\(AO\) and \(BO\) with \(AO\) and \(DO\))
• Fourth blank: angle (\(\angle AOB\) and \(\angle AOD\))
• Fifth blank: \(AOD\)
• Sixth blank: \(AD\)

Problem 4 (Proving \(BE = DF\) in rhombus \(ABCD\))

Step1: Identify the right angle

\(AE\perp BC\) and \(AF\perp CD\), so \(\angle AEB=\angle AFD = 90^\circ\) (since \(AE\) and \(AF\) are perpendiculars to \(BC\) and \(CD\) respectively).

Step2: Angles in a rhombus

In rhombus \(ABCD\), \(AB = AD\) (all sides of a rhombus are equal) and \(\angle ABE=\angle ADF\) (opposite angles of a rhombus are equal, and also \(BC\parallel AD\), \(AB\parallel CD\), so corresponding angles are equal).

Step3: Congruence of triangles

We have \(\angle AEB=\angle AFD = 90^\circ\) (step 1), \(\angle ABE=\angle ADF\) (step 2), and \(AB = AD\) (side of rhombus). By AAS (Angle - Angle - Side) congruence criterion, \(\triangle ABE\cong\triangle ADF\) (two angles and a non - included side are equal: \(\angle AEB=\angle AFD\), \(\angle ABE=\angle ADF\), \(AB = AD\)).

Step4: Equal segments from congruence

Since \(\triangle ABE\cong\triangle ADF\), corresponding sides \(BE = DF\).

Filling the blanks:

• First blank: \(AFD\)
• Second blank: \(AD\)
• Third blank: angles (\(\angle AEB\) and \(\angle AFD\), \(\angle ABE\) and \(\angle ADF\))
• Fourth blank: side (\(AB = AD\))
• Fifth blank: \(ABE\)
• Sixth blank: \(ADF\)
• Seventh blank: \(DF\)

Answer:

s:

Problem 3 Blanks:

1. \(AOD\)
2. \(DO\)
3. sides
4. angle
5. \(AOD\)
6. \(AD\)

Problem 4 Blanks:

1. \(AFD\)
2. \(AD\)
3. angles
4. side
5. \(ABE\)
6. \(ADF\)
7. \(DF\)