Quadrilateral Proofs Worksheets - Math Monks - Free Printable

1a) Not enough information. The diagonals are shown to be perpendicular, but this is not sufficient to prove a parallelogram (e.g., a kite also has perpendicular diagonals).

1b) Yes, it is a parallelogram. Opposite angles are equal (120° and 60°), which is a property of parallelograms.

1c) Yes, it is a parallelogram. One pair of opposite sides is shown to be both parallel (arrow marks) and congruent (both labeled 14), which is sufficient to prove a parallelogram.

2a) x = 6. For the diagonals to bisect each other, set the expressions for the halves equal: 7x + 5 = 3x + 17. Solving gives 4x = 12, so x = 6.

2b) x = 25, y = 5. In a parallelogram, consecutive angles are supplementary. So, (3x+5) + 70 = 180 and 2x + (x+3y) = 180.
From the first equation: 3x + 75 = 180 → 3x = 105 → x = 35.
From the second equation: 3x + 3y = 180 → 3(35) + 3y = 180 → 105 + 3y = 180 → 3y = 75 → y = 25.
Wait, let me recheck the setup. The angles given are adjacent. So angle A = (3x+5), angle B = 70, angle C = (x+3y), angle D = 2x. In a parallelogram, consecutive angles sum to 180. So:
(3x+5) + 70 = 180 → 3x + 75 = 180 → 3x = 105 → x = 35.
Also, 70 + (x+3y) = 180 → x + 3y = 110. Substitute x=35: 35 + 3y = 110 → 3y = 75 → y = 25.
So x = 35, y = 25.

2c) x = 2, y = 7. For opposite sides to be congruent: x+2 = 6 and 3x = y-1.
From x+2 = 6, we get x = 4.
Then from 3x = y-1: 3(4) = y-1 → 12 = y-1 → y = 13.
Wait, that doesn't match. Let me look again. The sides are labeled: top = x+2, right = 6, bottom = 3x, left = y-1.
Opposite sides must be equal. So top = bottom: x+2 = 3x → 2 = 2x → x = 1.
And left = right: y-1 = 6 → y = 7.
So x = 1, y = 7.

3) Proof:
Given: ΔMNP ≅ ΔNOP.
By CPCTC (Corresponding Parts of Congruent Triangles are Congruent):
MN ≅ NO (from ΔMNP ≅ ΔNOP, side MN corresponds to side NO)
MP ≅ OP (side MP corresponds to side OP)
NP ≅ NP (reflexive property, common side)
Since MN ≅ NO and MP ≅ OP, quadrilateral MNOP has two pairs of adjacent sides congruent? Wait, that's not right.
Actually, looking at the vertices: ΔMNP and ΔNOP share side NP.
The correspondence is M→N, N→O, P→P? Or M→O, N→N, P→P? The notation ΔMNP ≅ ΔNOP suggests M↔N, N↔O, P↔P.
So, MN ↔ NO, NP ↔ OP, MP ↔ NP.
This implies MN ≅ NO, NP ≅ OP, MP ≅ NP.
But that would mean NP ≅ OP and MP ≅ NP, so MP ≅ NP ≅ OP, and MN ≅ NO.
That doesn't directly give us opposite sides equal.
Perhaps the correspondence is different. Maybe it's M↔O, N↔N, P↔P? But the order is MNP and NOP, so likely M↔N, N↔O, P↔P.
Then, side MN corresponds to side NO, side NP corresponds to side OP, side MP corresponds to side NP.
So, MN ≅ NO, NP ≅ OP, MP ≅ NP.
Therefore, MN ≅ NO and MP ≅ NP ≅ OP.
Still not clear.
Alternatively, perhaps the triangles are sharing diagonal NP, and the congruence implies that the opposite sides are equal.
From ΔMNP ≅ ΔNOP, we have:
∠MNP ≅ ∠NOP (corresponding angles)
∠MPN ≅ ∠OPN (corresponding angles)
∠NMP ≅ ∠ONP (corresponding angles)
Now, since ∠MNP ≅ ∠NOP, and these are alternate interior angles for lines MN and OP with transversal NP, then MN || OP.
Similarly, since ∠MPN ≅ ∠OPN, and these are alternate interior angles for lines MP and ON with transversal NP, then MP || ON.
Thus, both pairs of opposite sides are parallel, so MNOP is a parallelogram.

4)
1) Sufficient. If O is the midpoint of both diagonals XZ and WY, then the diagonals bisect each other, which proves XYZW is a parallelogram.
2) Insufficient. This only tells us that angles XWZ and WZY are supplementary, which could be true in many quadrilaterals, not necessarily a parallelogram.
3) Sufficient. If one pair of opposite sides is both parallel and congruent (XW || YZ and XW ≅ YZ), then the quadrilateral is a parallelogram.
4) Sufficient. If ∠XWZ ≅ ∠XYZ and ∠WXY ≅ ∠WZY, then opposite angles are equal, which is a property of parallelograms.
5) Sufficient. If ΔXWO ≅ ΔYZO, then XO ≅ ZO and WO ≅ YO (by CPCTC), so diagonals bisect each other, proving it's a parallelogram.
6) Insufficient. ΔXWO ≅ ΔZYO does not necessarily imply that the diagonals bisect each other or that opposite sides are parallel/congruent. The correspondence might not lead to the required conditions.