In figure, AD ⊥ CD and CB ⊥ CD.

Given in the figure, AD ⊥ CD and CB ⊥ CD.

And AQ = BP and DP = CQ,

To prove that ∠DAQ = ∠CBP

Given that DP = QC

Add PQ on both sides

DP + PQ = PQ + QC

DQ = PC ... (i)

Now, consider triangle DAQ and CBP,

We have

∠ADQ = ∠BCP = 90°   [given] AQ = BP [given]

And DQ = PC [From (i)]

So, by RHS congruence criterion, we have

ΔDAQ ≅ ΔCBP

Now,

∠DAQ = ∠CBP   [Corresponding parts of congruent triangles are equal]

Hence proved