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Solution:
(i) From the figure △ AOB and △ DOC
We know that AB || CD and ∠ BAO and ∠ CDO are alternate angles
So we get∠ BAO = ∠ CDO
From the figure we also know that O is the midpoint of the line AD
We can write it as AO = DO
According to the figure we know that ∠ AOB and ∠ DOC are vertically opposite
angles.So we get ∠ AOB = ∠ DOC
Therefore, by ASA congruence criterion we get
△ AOB ≅△ DOC
(ii) We know that △ AOB ≅△ DOC
So we can write it as
BO = CO (c. p. c. t)
Therefore, it is proved that O is the midpoint of BC.

2.In the given figure, AD and BC are equal perpendiculars to a line segment AB.
Show that CDbisects AB.

Solution: Based on the △ AOD and △ BOC
From the figure we know that ∠ AOD and ∠ BOC are vertically opposite angles.
So we get
∠ AOD = ∠ BOC
We also know that
o
∠ DAO = ∠ CBO = 90
It is given that AD = BC
Therefore, by AAS congruence criterion we get
△ AOD ≅△ BOC
So we get AO = BO (c. p. c. t)
Therefore, it is proved that CD bisects AB.

3.In the given figure, two parallel lines l and m are intersected by two parallel
lines p and q. Show that △ABC ≅△CDA.
Solution: Based on the △ ABC and △ CDA
We know that p || q and ∠ BAC and ∠ DCA are alternate interior angles
So we get∠ BAC = ∠ DCA
We know that A and C are the common points for all the lines
So it can be written asAC = CA
We know that l || m and ∠ BCA and ∠ DAC are alternate interior angles
So we get∠ BCA = ∠ DAC
Therefore, by ASA congruence rule it is proved that
△ ABC ≅△ CDA
4. In the given figure, OA = OB and OP = OQ. Prove that
(i) PX = QX,