In the figure AD=BD=CD .FIND ANGEL BAC

Given that if AD=BD=CD

we have to find the ∠BAC

In ΔADC, AD=DC

⇒ ADC is an isosceles triangle

⇒ ∠DAC=∠DCA   →    (1)

Similarly, In ΔADB, AD=DB

⇒ ADB is an isosceles triangle

⇒ ∠DAB=∠DBA   →    (2)

In ΔABC, by angle sum property

∠DBA+∠DCA+∠BAC=180°

∠DBA+∠DCA+∠DAC+∠DAB=180°

∠DBA+2∠DAC+∠DAB=180°    (Using (1))

2∠DAB+2∠DAC=180°              (Using (2))

∠DAB+∠DAC=90°

∠BAC=90°