Question

please answer thanks :)​

Answer 1

Answer:

Here we are given with a figure which is made up of two triangles intersecting each other and we have to prove the below conditions asked in the Question$.$

GiveN:

• AB = DC
• ∠ABC = ∠DCB

To prove:

• ∆ABC ≅ ∆DBC
• ∠A = ∠D
• ∆AOB ≅ ∆DOC

Proof:

(1) In ∆ABC and ∆DBC,

• AB = DC (given)
• BC = BC (Common)
• ∠ABC = ∠DCB

Hence, ∆ABC ≅ ∆DBC (SAS congurence)

(2) Since from (1),

• ∆ABC ≅ ∆DBC

Then,

➝ ∠A = ∠D (CPCT)

This is because ∆ABC is congurent to ∆DBC, and hence ∠A is equals to ∠D because they are the corresponding angles.

(3) In ∆AOB and ∆DOC,

• ∠A = ∠D (proved above)
• ∠AOB = ∠DOC (Vertically opposite angles)
• AB = DC (given)

Hence, ∆AOB ≅ ∆DOC (AAS congurence)

And we are done proving them !!

Answer 2

a) in the triangle ABC and DBC

AB=DC (given)

angle ABC =angle DCB (given)

BC=BC (common side)

so, triangle ABC ~ triangle DBC

b) if tringle ABC~DBC so

angle A=angle D

c) in triangle AOB and DOC

AB=DC (given)

angle A = angle D (proved)

OB=OC (equal angle ke samne wali sides bhi equal hoti hai)

so

triangle AOB~DOC

I am 100% sure that is is correct answer so PLZ mark me as a brainlist