Question

In the given figure AD is perpendicular to CD and BC is perpendicular to CD. If AQ = BP and DP=CQ, show 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:

Step-by-step explanation:

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.