Given: m∠adb = m∠cdb ad ≅ dc prove: m∠bac = m∠bca bd⊥ ac

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

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step-by-step explanation: