Question

In the figure BD = DC and angle DBC = 25. Find the measure of BAC.

Answer 1

Answer:

In ∆BCD, we have

$\\ \qquad \sf \: BD = DC \\ \\ \\ \implies \sf \angle \: DCB = \angle \: DBC \\ \\ \\ \implies \sf \: \angle \: DCB = 25 {}^{ \circ}$

Also,

$\\ \sf \angle \: DCB + \angle \: DBC + \angle \: BDC = 180 \\ \\ \\ \implies \sf \: 25 + 25 + \angle \: BDC = 180 \\ \\ \\ \implies \sf \angle \: BDC = 130 {}^{ \circ}$

Since ABCD is a cyclic quadrilateral.

$\\ \therefore \: \: \: \: \sf \angle \: BAC + \angle \: BDC = 180 \\ \\ \\ \implies \sf \angle \: BAC+ 130 = 180 \\ \\ \\ \implies \sf \: \angle \: BAC = 50 {}^{ \circ}$