Question

3 in parallelogram abcd on the right, given the length of the diagonals are equal, prove that it is a rectangle as follows, fill the appropriate words in the blanks.
way of thinking
to prove that abcd is a rectangle, use the definition that the four angles of a rectangle are
proof
from parallelogram abcd, ( ac = bd )
in ( \triangle abc ) and ( \triangle dcb ),
since the opposite sides of the parallelogram are equal,
( ab=)
given ( ac=)
also, is shared
from ①, ②, and ③, since the three are equal respectively,
( \triangle abccong\triangle dcb ).
therefore, ( angle abc=angle )
since quadrilateral abcd is a parallelogram,
( angle abc=angle adc ) and ( angle dcb=angle )
from ④ and ⑤, since the four are equal,
parallelogram abcd is a rectangle.

Explanation:

Step1: Recall rectangle angle definition

A rectangle is defined as a parallelogram with four right angles (or four angles equal to \(90^\circ\)). So the first blank (for the angle definition) is "right angles (or \(90^\circ\) angles)".

Step2: Opposite sides of parallelogram

In a parallelogram, opposite sides are equal. So in parallelogram \(ABCD\), \(AB = DC\) (since \(AB\) and \(DC\) are opposite sides). So the first blank in the proof (①) is \(DC\).

Step3: Given diagonal equality

The problem states that the diagonals of the parallelogram are equal, so \(AC = BD\) (given). So the second blank (②) is \(BD\).

Step4: Shared side in triangles

In \(\triangle ABC\) and \(\triangle DCB\), the side \(BC\) is shared (common side). So the third blank (③) is \(BC\).

Step5: Corresponding sides for congruence

For triangle congruence (SSS, since \(AB = DC\), \(AC = BD\), \(BC = BC\)), the three "sides" are equal respectively. So the fourth blank (after "three") is "sides".

Step6: Corresponding angles from congruence

Since \(\triangle ABC \cong \triangle DCB\), corresponding angles \(\angle ABC=\angle DCB\)? Wait, no, wait: Wait, \(\triangle ABC\) and \(\triangle DCB\) – let's correct. Wait, in parallelogram \(ABCD\), \(AB \parallel DC\) and \(AD \parallel BC\). Wait, maybe I mixed up the triangles. Wait, actually, in \(\triangle ABC\) and \(\triangle BAD\)? No, the problem has \(\triangle ABC\) and \(\triangle DCB\). Wait, \(AB = DC\), \(BC = CB\) (shared), \(AC = DB\) (given). So by SSS, \(\triangle ABC \cong \triangle DCB\). Then corresponding angles: \(\angle ABC=\angle DCB\). But since \(ABCD\) is a parallelogram, \(\angle ABC + \angle BCD = 180^\circ\) (consecutive angles in parallelogram are supplementary). If \(\angle ABC=\angle BCD\), then each is \(90^\circ\). Wait, maybe the triangles are \(\triangle ABC\) and \(\triangle BAD\)? No, the problem says \(\triangle ABC\) and \(\triangle DCB\). Wait, let's re - examine.

Wait, in the proof, after \(\triangle ABC \cong \triangle DCB\), then \(\angle ABC=\angle DCB\). But in a parallelogram, \(AB \parallel DC\), so \(\angle ABC + \angle DCB = 180^\circ\) (consecutive interior angles). If \(\angle ABC=\angle DCB\), then \(\angle ABC=\angle DCB = 90^\circ\). Then, since it's a parallelogram, all angles are \(90^\circ\). Wait, but the next step: "Therefore, \(\angle ABC=\angle\) [blank]". Wait, maybe I made a mistake in triangle labeling. Let's start over.

In parallelogram \(ABCD\), \(AB = DC\), \(BC = AD\), \(AC = BD\) (given). Consider \(\triangle ABC\) and \(\triangle DCB\): \(AB = DC\), \(BC = CB\), \(AC = DB\). So \(\triangle ABC \cong \triangle DCB\) (SSS). Therefore, \(\angle ABC=\angle DCB\). But in parallelogram \(ABCD\), \(AB \parallel DC\), so \(\angle ABC + \angle DCB = 180^\circ\) (consecutive angles). So \(\angle ABC=\angle DCB = 90^\circ\). Then, since \(ABCD\) is a parallelogram, \(\angle ABC=\angle ADC\) (opposite angles of parallelogram) and \(\angle DCB=\angle DAB\) (opposite angles). Wait, the problem says "Therefore, \(\angle ABC=\angle\) [blank]". Wait, maybe the triangles are \(\triangle ABC\) and \(\triangle BAD\)? No, the problem has \(\triangle ABC\) and \(\triangle DCB\). Wait, perhaps the correct corresponding angle is \(\angle DCB\)? Wait, no, let's check the next step: "Since quadrilateral \(ABCD\) is a parallelogram, \(\angle ABC + \angle BCD = 180^\circ\) (consecutive angles). If \(\angle ABC=\angle BCD\), then each is \(90^\circ\). Then, as a parallelogram with one right angle, it's a rectangle.

Wait, the fourth blank (after \(\angle ABC=\angle\)): from \(\triangle ABC \cong \triangle DCB\), \(\angle ABC=\angle DCB\). So the blank is \(\angle DCB\)? Wait, no, the problem's step ④: "Therefore, \(\angle ABC=\angle\) [blank]". Wait, maybe I messed up the triangle congruence. Let's consider \(\triangle ABC\) and \(\triangle BAD\): \(AB = BA\), \(BC = AD\), \(AC = BD\) (given). So \(\triangle ABC \cong \triangle BAD\) (SSS). Then \(\angle ABC=\angle BAD\). But in parallelogram, \(\angle ABC + \angle BAD = 180^\circ\), so each is \(90^\circ\). Hmm, maybe the problem has a typo, but following the standard proof: In a parallelogram, if diagonals are equal, then it's a rectangle. The proof uses SSS on \(\triangle ABC\) and \(\triangle DCB\) (or \(\triangle ABC\) and \(\triangle BAD\)) to show that a pair of consecutive angles are equal and supplementary, hence right angles.

Continuing:

Step7: Consecutive angles in parallelogram

Since \(ABCD\) is a parallelogram, \(\angle ABC + \angle BCD = 180^\circ\) (consecutive angles). But from congruence, \(\angle ABC=\angle BCD\), so each is \(90^\circ\). Then, opposite angles of parallelogram are equal, so \(\angle ABC=\angle ADC\) and \(\angle BCD=\angle DAB\), so all angles are \(90^\circ\).

Wait, the fifth blank (after \(\angle DCB=\angle\)): \(\angle DAB\) (since \(\angle DCB\) and \(\angle DAB\) are opposite angles of parallelogram).

Step8: Four angles conclusion

Since the four "angles" are equal (all \(90^\circ\)), the parallelogram is a rectangle. So the last blank (after "four") is "angles".

Answer:

1. (First blank - angle definition): right angles (or \(90^\circ\) angles)
2. (①): \(DC\)
3. (②): \(BD\)
4. (③): \(BC\)
5. (After "three"): sides
6. (After \(\angle ABC=\angle\)): \(\angle DCB\) (or correct corresponding angle from congruence)
7. (After \(\angle DCB=\angle\)): \(\angle DAB\)
8. (After "four"): angles

(Note: The exact blanks may need to be adjusted based on the precise triangle labeling, but the key concepts are as above. The main blanks are filled with the logical geometric properties: opposite sides of parallelogram, equal diagonals, shared side, SSS congruence, corresponding angles, supplementary consecutive angles in parallelogram, opposite angles of parallelogram, and four right angles for rectangle.)