As shown in the right figure, in the parallelogram ABCD, we know that ad = 8 and perimeter is equal to 24, then we find the lengths of the other three sides

As shown in the right figure, in the parallelogram ABCD, we know that ad = 8 and perimeter is equal to 24, then we find the lengths of the other three sides

BC=AD=8,AB=CD=(24-2AD)/2=(24-2*8)/2=4
∴BC=8,
AB=4,
CD=4

As shown in the figure, the quadrilateral ABCD is a rectangle, the △ PBC and △ QCD are equilateral triangles, and the point P is above the rectangle, and the point q is inside the rectangle

It is proved that: (1) the ∵ quadrilateral ABCD is a rectangle. ∵ ABC = ∠ BCD = 90 ° (1 minute) ∵ the ∵ PBC and △ QCD are equilateral triangles. ∵ PBC = ∠ PCB = ∠ QCD = 60 ° (1 minute) ∵ PBA = ∠ ABC - ∠ PBC = 30 °, (1 minute) ∵ PCD = ∠ BCD - ∠ PCB = 30 °. ∵ PCQ = QCD - ∠ PCD = 30 °. ∵ PBA = ∠ PCQ = 30 ° (1 minute) (2) ∵ AB = DC = QC, ∵ PBA = PCQ, Pb = PC. (1 minute) ）≌△ PAB ≌△ PQC. (2 points) ≌ PA = PQ. (1 point)

As shown in the figure, the quadrilateral ABCD is a rectangle, and △ PBC and △ QCD are equilateral triangles
How do I find PAB and PQC congruent & nbsp;, how do I feel less condition

Certification:
∵ the quadrilateral ABCD is a rectangle
∴AB=CD,∠ABC=∠BCD=90º
Both ⊿ PBC and ⊿ QCD are equilateral triangles
∴PB=PC,CQ=CD=AB,∠PBC=∠PCB=∠QCD=60º
∵∠PBA=∠ABC+∠PBC=90º+60º=150º
∠PQC=360º-∠QCD-∠BCD-∠PCB=360º-60º-90º-60º=150º
∴∠PBA=∠PQC.（1）
And ∵ AB = CQ, Pb = PC
∴⊿PBA≌⊿PCQ（SAS）
∴PA=PQ.（2）

As shown in the figure, the quadrilateral ABCD is a rectangle, the △ PBC and △ QCD are equilateral triangles, and the point P is above the rectangle, and the point q is inside the rectangle

It is proved that: (1) the ∵ quadrilateral ABCD is a rectangle, ABC = ∠ BCD = 90 ° (1 minute) ∵ △ PBC and △ QCD are equilateral triangles, PBC = ∠ PCB = ∠ QCD = 60 ° (1 minute) ∵ PBA = ∠ ABC - ∠ PBC = 30 °, (1 minute) ∵ PCD = ∠ BCD - ∠ PCB = 30 °, PCQ = ∠ QCD - ∠ PCD = 30 °