In the given figure, ΔABD is an isosceles triangle with AB=AD. A median is drawn from A to side BD at C. Which of the following option(s) is/are correct?
AC is perpendicular to BD
∠BAC=∠DAC
ΔABC≅ΔADC
Area of ΔABD=12×Area of ΔABC
Consider ΔABC and ΔADC,AC=AC (Common side of both triangles)AB=AD (Sides of isosceles triangle)BC=CD (Median divides a side in two equal parts)∴ΔABC≅ΔADC (by SSS congruence criterion) So the option "ΔABC≅ΔADC" is correct.∴∠ACB=∠ACD (CPCT)But they are supplementary angles. So, ∠ACB+∠ACD=180∘⇒∠ACB=∠ACD=90∘∴AC is perpendicular to BD and the hence option "AC is perpendicular to BD" is correct. Also, ∠BAC=∠DAC (corresponding angles of congruent triangles)Hence, the option"∠BAC=∠DAC" is also correct.Also, area of ΔABC=area of ΔADC=12(area of ΔABD).ΔABC≅ΔADC⟹ ar(ΔABC)=ar(ΔADC)Hence, the option "Area of ΔABD=12×Area of ΔABC"is also correct.