Question

An isosceles triangle ABC has AC = BC. CD bisects AB at D and $\angle CAB={55}^{\circ }$. Find $\angle DCB$.

• A. ${32}^o$
• B. ${42}^o$
• C. ${35}^o$
• D. none of the above

Answer 1

The correct option is C ${35}^o$

Draw a diagram with given condition,

Given $AC=BC$
thus, $\angle ABC=\angle BAC={55}^{\circ }$ (Isosceles triangle property)
Now, in $△$s CDA and CDB
$AC=BC$ (Given)
$BD=AD$ (CD bisects AB)
$CD=CD$ (Common side)
Hence. $△CDA\cong △CDB$
thus, $\angle BCD=\angle ACD=x^{\circ }$ (Corresponding angles of congruent triangles)
Now, In $△ABC$,
$\angle ABC+\angle ACB+\angle BAC={180}^{\circ }$
${55}^{\circ }+\angle ACD+\angle BCD+{55}^{\circ }={180}^{\circ }$
$2x^{\circ }={70}^{\circ }$
$x^{\circ }={35}^{\circ }$
Thus, $\angle DCB={35}^{\circ }$