Given that in figure,
AD ⊥ CD and CB ⊥ CD
And,
AQ = BP, DP = CQ
We have to prove that,
∠DAQ = ∠CBP
Given that, DP = QC
Adding PQ on both sides, we get
DP + PQ = PQ + QC
DQ = PC (i)
Now consider, ΔDAQ and ΔCBP we have
∠ADQ = ∠BCP = 90°(Given)
AQ = BP (Given)
And,
DQ = PC (From i)
So, by RHS congruence rule, we have
ΔDAQ ≅ ΔCBP
Now,
∠DAQ = ∠CBP (By c.p.c.t)
Hence, proved