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 ___.

• A.
• B.
• C.
• D. Can not be determined

Answer 1

The correct option is A

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)

In $△$BCD,

$\angle$DBC + $\angle$BCD + $\angle$CDB= ${180}^{\circ }$

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

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

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

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

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