Question

In the given figure, seg AD perpendicular to seg CD and
seg BC perpendicular to seg CD. If DP = CQ and AP = BQ.
then prove that (triangle)DAQ congruent to triangle CBP​

Answer 1

Step-by-step explanation:

Given that, in the figure AD⊥CD and CB⊥CD. AQ = BP and DP = CQ

Figure 1

We have to prove that ∠DAQ=∠CBP

Now, consider triangle DAQ and CBP,

figure 2

We have So, by RHS congruence criterion, we have ΔDAQ≅ΔCBP

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

∴ Hence proved

Answer 2

Answer:

angle DAQ= angle CBP (Corresponding part of congruent triangles)

Step-by-step explanation:

Solution:

Given AD 1 CD and BCL CD AQ = BP and DP = CQ

To prove: angle DAQ= angle CBP

Proof:

AD1 CD and BC 1 CD

:: angle D= angle C (each 90°)

:: DP = CQ (Given)

Adding PQ to both sides. we get DP + PQ = PQ + CQ

Rightarrow DQ+CP

Now, in right angles ADQ and BPC

:. Hyp. AQ = Hyp. BP

Side DQ = side CP

:. Delta ADQ equiv Delta BPC (Right angle hypotenuse side)

:. angle DAQ= angle CBP (Corresponding part of congruent triangles)

Hence proved.