Question

Maths 2 Similarity, solve this. as soon as possible
Spam = Report​

Answer 1

• Answer:
• In ΔABC and ΔPQR
• In ΔABC and ΔPQRAB=PQ
• In ΔABC and ΔPQRAB=PQAnd BC=QR
• In ΔABC and ΔPQRAB=PQAnd BC=QRCB And RQ extended  to X and Y
• In ΔABC and ΔPQRAB=PQAnd BC=QRCB And RQ extended  to X and Y∠ABX=∠PQY
• In ΔABC and ΔPQRAB=PQAnd BC=QRCB And RQ extended  to X and Y∠ABX=∠PQY∠ABC=180o−∠ABX=180o∠PQY
• In ΔABC and ΔPQRAB=PQAnd BC=QRCB And RQ extended  to X and Y∠ABX=∠PQY∠ABC=180o−∠ABX=180o∠PQY∠PQR=180o−∠PQY=∠ABC
• In ΔABC and ΔPQRAB=PQAnd BC=QRCB And RQ extended  to X and Y∠ABX=∠PQY∠ABC=180o−∠ABX=180o∠PQY∠PQR=180o−∠PQY=∠ABCWe get in both triangles 
• In ΔABC and ΔPQRAB=PQAnd BC=QRCB And RQ extended  to X and Y∠ABX=∠PQY∠ABC=180o−∠ABX=180o∠PQY∠PQR=180o−∠PQY=∠ABCWe get in both triangles AB=PQ
• In ΔABC and ΔPQRAB=PQAnd BC=QRCB And RQ extended  to X and Y∠ABX=∠PQY∠ABC=180o−∠ABX=180o∠PQY∠PQR=180o−∠PQY=∠ABCWe get in both triangles AB=PQBC=QR
• In ΔABC and ΔPQRAB=PQAnd BC=QRCB And RQ extended  to X and Y∠ABX=∠PQY∠ABC=180o−∠ABX=180o∠PQY∠PQR=180o−∠PQY=∠ABCWe get in both triangles AB=PQBC=QRAnd ∠ABC=∠PQR
• In ΔABC and ΔPQRAB=PQAnd BC=QRCB And RQ extended  to X and Y∠ABX=∠PQY∠ABC=180o−∠ABX=180o∠PQY∠PQR=180o−∠PQY=∠ABCWe get in both triangles AB=PQBC=QRAnd ∠ABC=∠PQRThen both triangle are congruent
• In ΔABC and ΔPQRAB=PQAnd BC=QRCB And RQ extended  to X and Y∠ABX=∠PQY∠ABC=180o−∠ABX=180o∠PQY∠PQR=180o−∠PQY=∠ABCWe get in both triangles AB=PQBC=QRAnd ∠ABC=∠PQRThen both triangle are congruenti.e. ΔABC=ΔPQR

Step-by-step explanation:

In ΔABC and ΔPQR

AB=PQ

And BC=QR

CB And RQ extended  to X and Y

∠ABX=∠PQY

∠ABC=180o−∠ABX=180o∠PQY

∠PQR=180o−∠PQY=∠ABC

We get in both triangles 

AB=PQ

BC=QR

And ∠ABC=∠PQR

Then both triangle are congruent

i.e. ΔABC=ΔPQR

Answer 2

Answer:

B] 1. ABC = QPR

It is 1 mark question.

May it helps you.