Question

in the figure, ad is perpendicular to cd and bc is perpendicular to cd. if aq=bp and dp=cq. prove that angle daq = angle cbp

Answer 1

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.

Answer 2

Answer:

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.

Step-by-step explanation: