It is known that the perimeter of the parallelogram ABCD is 32, ab = 4, then BC is equal to

Because it is a parallelogram, so AB = CD; BC = ad.. 32-ab-cd = BC + ad. because AB = CD; BC = ad. so 32-4-4 = BC + BC. Find BC = 12

In the parallelogram ABCD, if AB = 7 and the circumference is equal to 20, then BC=

AB = CD ad = BC
Perimeter: 2 (AB + BC) = 20
That is 7 + BC = 10
∴BC=3

In the parallelogram ABCD, points m and N are the midpoint of AB and CD respectively, and BD intersects an and cm at points P and Q respectively. It is proved that DP = PQ = QB
In the parallelogram ABCD, points m and N are the midpoint of AB and CD respectively, and BD intersects an and cm at points P and Q respectively. It is proved that DP = PQ = QB,
You can prove that it is a median line on one condition. How can you prove it? To prove the median line, you must have PQ = QB

Because points m and N are the midpoint of AB and CD respectively
So AB = CD, am = BM = CN = DN
And ∥ an ∥ cm
Ψ PN is the median of Δ DQC and QM is the median of Δ BPA,
∴DP=PQ,PQ=QB
DP=PQ=QB
The proof is complete

In the parallelogram ABCD, the diagonal lines AC and BD intersect at o m n, which are the midpoint of OA OC respectively. What is the relationship between BM and DN?

Prove: connect DM and BN, prove triangle (omit later) OMB and ond congruence (corner edge) to get BM = DN, prove ODM and OBN congruence (corner edge) to get DM = BN, then prove quadrilateral bndm is parallelogram to get BM parallel DN. In conclusion, BM is parallel and equal to dn, listen to the rain, Mint has no Chen. PS: think more about yourself, Come on! You can write it if you think about it,

It is known that the perimeter of the parallelogram ABCD is 32, ab = 4, then BC is equal to