$\text{Consider}\Delta ABC\text{and}\Delta ADC,AC=AC\text{(Common side of both triangles)}AB=AD\text{(Sides of isosceles triangle)}BC=CD\text{(Median divides a side in two equal parts)}\therefore \Delta ABC\cong \Delta ADC\left(\text{by SSS congruence criterion}\right)\text{So the option}"\Delta ABC\cong \Delta ADC"\text{is correct.}\therefore \angle ACB=\angle ACD\left(CPCT\right)\text{But they are supplementary angles. So,}\angle ACB+\angle ACD={180}^{\circ }\Rightarrow \angle ACB=\angle ACD={90}^{\circ }\therefore AC\text{is perpendicular to}BD\text{and the hence option "AC is perpendicular to BD" is correct.}\text{Also,}\angle BAC=\angle DAC\text{(corresponding angles of congruent triangles)}\text{Hence, the option}"\angle BAC=\angle DAC"\text{is also correct.}\text{Also, area of}\Delta ABC=\text{area of}\Delta ADC=\frac{1}{2}\left(\text{area of}\Delta ABD\right).\Delta ABC\cong \Delta ADC⟹ar\left(\Delta ABC\right)=ar\left(\Delta ADC\right)\text{Hence, the option "Area of}\Delta ABD=\frac{1}{2}\times \text{Area of}\Delta ABC"\text{is also correct.}$