Question

In Fig. 10.99, AD ⊥ CD and CB ⊥ CD. If AQ=BP and DP =CQ, prove that ∠DAQ = ∠CBP.

Answer 1

Concept :  

Congruence of triangles:

• Two ∆’s are congruent if sides and angles of a triangle are equal to the corresponding sides and angles of the other ∆.
•  In Congruent Triangles corresponding parts are always equal and we write it in short CPCT i e, corresponding parts of Congruent Triangles.
• It is necessary to write a correspondence of vertices correctly for writing the congruence of triangles in symbolic form.

   

 

Given : AD ⊥ CD and CB ⊥ CD and AQ = BP , DP =CQ.

To Prove : ∠DAQ = ∠CBP.

Proof:

We have DP = QC

On adding PQ on both sides, we obtain :  

DP + PQ = PQ + QC

DQ = PC ………... (1)

 

Now,

In ∆DAQ and ∆CBP , we have

∠ADQ = ∠BCP = 90° (Given)

AQ = BP (Given)

DQ = PC (From eq 1)

∆DAQ ≅ ∆CBP

[By RHS congruence criterion]

∠DAQ = ∠CBP (By c.p.c.t)

Hence, proved

HOPE THIS ANSWER WILL HELP YOU…..

 

 

Some questions of this chapter :

In Fig. 10.92, it is given that AB = CD and AD = BC. Prove that Δ ADC ≅ Δ CBA.

https://brainly.in/question/17044575

AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (See the given figure). Show that (i) ΔDAP ≅ ΔEBP (ii) AD = BE

https://brainly.in/question/1399425

Answer 2

Step-by-step explanation:

In ∆DAQ and ∆CBP , we have

∠ADQ = ∠BCP = 90° (Given)

AQ = BP (Given)

DQ = PC (From eq 1)

∆DAQ ≅ ∆CBP

[By RHS congruence criterion]

∠DAQ = ∠CBP (By c.p.c.t)

Hence, proved

HOPE THIS ANSWER WILL HELP YOU…..