In the given figure, ABCD is a rectangle and X and Y are the mid-points of the sides DC and AB respectively. P and Q are the points of AD and BC respectively such that DP = BQ.

Prove that △ APX ≅ △ CQY.

Given, ABCD is a rectangle.

X and Y are midpoints of DC and AB respectively.

P and Q are points on AD and BC respectively such that DP = BQ

Comparing sides AX and CY,

In rectangle, opposite sides are equal, so AB = CD.

Since X and Y are mid-points:

DX = $\dfrac{1}{2}CD = \dfrac{1}{2}AB$

YB = $\dfrac{1}{2}AB$

Therefore, DX = YB……..(1)

Consider △ ADX and △ CBY,

⇒ AD = CB (opposite sides of a rectangle)

⇒ ∠D = ∠B = 90°

⇒ DX = YB (Proven above)

So, △ ADX ≅ △ CBY (By S.A.S axiom)

∴ AX = CY (By C.P.C.T.C.) ……..(2)

Comparing sides AP and CQ,

Given,

AD = BC and DP = BQ.

AD - DP = BC - BQ

∴ AP = CQ …….(3)

Consider △ DPX and △ BQY:

⇒ DP = BQ (given)

⇒ ∠D = ∠B = 90°

⇒ DX = BY (Proved above)

So, △ DPX ≅ △ BQY (By S.A.S axiom)

∴ PX = QY ……….(4)

Consider △ APX and △ CQY.

⇒ AX = CY (From equation (2))

⇒ AP = CQ (From equation (3))

⇒ PX = QY (From equation (4))

Since all the three corresponding sides are equal,

∴ △ APX ≅ △ CQY (By S.S.S axiom)

Hence proved that s△ APX ≅ △ CQY.