Question

$△ABC$ is an isosceles triangle, in which AB=AC.
Side BA is produced to D such that AD=AB.
$\angle BCD$ is equal to __${}^{\circ }$.

• A. ${90}^{\circ }$
• B. ${45}^{\circ }$
• C. ${37}^{\circ }$
• D. $\text{Cannot be determined}$

Answer 1

The correct option is A

${90}^{\circ }$

Given, AB=AC and BA=AD

In $△ABC,$

$\angle ABC=\angle ACB=x$
(opposite angles of equal sides)

In $△ACD,$

$\angle ADC=\angle ACD=y$
(opposite angles of equal sides)

$\angle BCD=x+y$

$In△ABC,$

$\angle ABC+\angle ACB+\angle BAC={180}^{\circ }$

$x+x+\angle BAC={180}^{\circ }$

$\angle BAC=180-2x$

$\angle BAC+\angle DAC={180}^{\circ }$
$(pointsB,AandDarecollinear)$

${180}^{\circ }-2x+\angle DAC={180}^{\circ }$

$So,\angle DAC=2x$

$In△DAC,$

$\angle DAC+\angle ADC+\angle ACD={180}^{\circ }$

$2x+y+y={180}^{\circ }$

$2x+2y={180}^{\circ }$

$x+y={90}^{\circ }$

$\angle BCD=\angle BCA+\angle ACD=(x+y)={90}^{\circ }$

$\angle BCD={90}^{\circ }$