Question

Given: m∠ADB = m∠CDB

AD ≅ DC
Prove: m∠BAC = m∠BCA
BD⊥ AC

Answer 1

Answer: Nah, it isn't ASA. Its SAS.

measure of ADB = measure of CBD by SAS(Side-Angle-Side)theorem.

Answer 2

The proof is explained below.

Explanation:

Given m∠ADB = m∠CDB and AD ≅ DC

we have to prove that m∠BAC = m∠BCA and BD⊥ AC

In ΔADO and ΔCDO

∠OAD=∠OCD (∵ADC is an isosceles triangle)

AD=DC (∵Given)

∠ADO=∠CDO (∵Given)

By ASA rule, ΔADO≅ΔCDO

In ΔBAD and ΔBCD

AD=DC (∵ABC is an isosceles triangle)

∠ADB=∠CDB (∵Given)

DB=DB (∵common)

By ASA rule, ΔADB≅ΔCDB

Now, ΔADB≅ΔCDB and ΔADO≅ΔCDO

⇒ ΔADB-ΔADO≅ΔCDB-ΔCDO

⇒ ΔABO≅ΔCBO

Hence, by CPCT, m∠BAC = m∠BCA

Now, we have to prove that BD⊥ AC i.e we have to prove m∠BOA=90°

Now, ΔABO≅ΔCBO therefore by CPCT, m∠BOA = m∠BOC

But, m∠BOA + m∠BOC=180° (linear pair)

⇒ m∠BOA + m∠BOA=180°

⇒ 2m∠BOA=180° ⇒ m∠BOA=90°

Hence, BD⊥ AC