Question

$\text{In the given figure,}\Delta ABD\text{is an isosceles triangle with AB=AD.}\text{A median is drawn from 'A'}\text{to side BD at C.}\text{Which of the following option(s) is/are correct?}$

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

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

$\text{Area of}\Delta ABC=\text{half of Area of}\Delta ABD$

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