Question

In the given figure, $\Delta ABD$ 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?

• A. $AC\text{is perpendicular to}BD$
• B. $\angle BAC=\angle DAC$
• C. $\Delta ABC\cong \Delta ADC$
• D. $\text{Area of}\Delta ABD=\text{half of}\Delta ABC$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $AC\text{is perpendicular to}BD$
B $\angle BAC=\angle DAC$
C $\Delta ABC\cong \Delta ADC$
D $\text{Area of}\Delta ABD=\text{half of}\Delta ABC$

Consider$\Delta ABC$ and $\Delta ADC$
AC = AC (Common side of both triangles)
AB = AD (Sides of isoscles triangle)
BC = CD (Median divides a side in two equal parts)
$\therefore \Delta ABC\cong \Delta ADC$.
$\therefore \angle ACB=\angle ACD$

But they are supplementary angles.
$\angle ACB+\angle ACD={180}^{\circ }$
$\Rightarrow \angle ACB=\angle ACD={90}^{\circ }$
$\therefore AC\perp BD$

Also$\angle BAC=\angle DAC$(corresponding angles of congruent triangles)

$\text{Also, area of}\Delta ABC=\text{area of}\Delta ADC=\frac{1}{2}\text{area of}\Delta ABD.$