Question

❗️Help needed please ❗️

Answer 1

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:

Answer 2

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