Answer:
The proof is explained below.
Step-by-step 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
