25. In the given figure AD is perpendicular to CD and BC is perpendicular to CD.
If AQ=BP and DP=CQ prove that ∆ DAQ~∆CBP
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Solution:-
Given : AD ⊥ CD and BC ⊥ CD
AQ = BP and DP = CQ
To prove : ∠ DAQ = ∠ CBP
Proof :
AD ⊥ CD and BC ⊥ CD
∴ ∠ D = ∠ C (each 90°)
∵ DP = CQ (Given)
Adding PQ to both sides. we get
DP + PQ = PQ + CQ
⇒ DQ + CP
Now, in right angles ADQ and BPC
∴ Hyp. AQ = Hyp. BP
Side DQ = side CP
∴ Δ ADQ ≡ Δ BPC (Right angle hypotenuse side)
∴ ∠ DAQ = ∠CBP (Corresponding part of congruent triangles)
Hence proved.
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