29.2k views
3 votes
❗️Help needed please ❗️

❗️Help needed please ❗️-example-1

2 Answers

2 votes

Answer:

AD cuts BC at I => I is the midpoint of BC => BI = CI

ΔAIB and ΔAIC have:

AB = AC

BI = CI

m∠ABC = m∠ACB ( because ΔABC is a isosceles triangle)

=> ΔAIB ≅ ΔAIC (sas)

=> m∠BAD = m∠CAD

ΔADB and ΔADC have:

m∠BAD = m∠CAD

AB = AC

AD is the common edge

=> ΔADB ≅ ΔADC (sas)

Explanation:

User Rodney Salcedo
by
5.7k points
4 votes

Answer:

prove it by S.S.S axiom bro

Explanation:

statement reason

1.in triangle ADB and ADC

a. AB=AC a.given

b.BD=CD b. radii of same circle

c. AD=AD c. common side

2. ∵triangle ADB is congurent to 2. by S.S.S axiom

ADC

User Kattie
by
5.2k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.