Question

line segment joining the midpoint M and N of parallel sides AB and DC respectively of a trapezium ABCD is perpendicular to both the sides AB and DC prove that AD= BC.​

Answer 1

Answer:

Construct AN and BN at the point N

Consider △ANM and ∠BNM

We know that N is the midpoint of the line AB

So we get

AM=BM

From the figure we know that

∠AMN=∠BMN=90

∘

MN is common i.e. MN=MN

By SAS congruence criterion

△ANM≅△BNM

AN=BN(c.p.c.t)…(1)

We know that

∠ANM=∠BNM(c.p.c.t)

Subtracting LHS and RHS by 90

∘

90

∘

−∠ANM=90

∘

−∠BNM

So we get

∠AND=∠BNC…(2)

Now, consider △AND and △BNC

AN=BN

∠AND=∠BNC

We know that N is the midpoint of the line DC

DN=CN

By SAS congruence criterion

△AND≅△BNC

AD=BC(c.p.c.t)

Therefore, it is proved that AD=BC.

Step-by-step explanation:

I hope it's helpful

Answer 2

Answer:

Line segment joining the midpoint M and N of parallel sides AB and DC respectively of a trapezium ABCD is perpendicular to both the sides AB and DC prove that AD= BC.

Step-by-step explanation:

${Sol}^{n} =$

Construct AN and BN at the point N

Consider △ANM and ∠BNM

We know that N is the midpoint of the line AB

So we get

AM = BM

From the figure we know that

∠AMN = ∠BMN = 90°

MN is common i.e. MN = MN

By SAS congruence criterion

△ANM ≅ △BNM

AN=BN(c.p.c.t)…(1)

We know that

∠ANM = ∠BNM(c.p.c.t)

Subtracting LHS and RHS by 90°

90° − ∠ANM = 90° − ∠BNM

So we get

∠AND = ∠BNC…(2)

Now, consider △AND and △BNC

AN = BN

∠AND = ∠BNC

We know that N is the midpoint of the line DC

DN = CN

By SAS congruence criterion

△AND ≅ △BNC

AD = BC(c.p.c.t)

Therefore, it is proved that AD=BC.