Question

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

Answer 1

In $△ABC$,

AC = BC (given)

$\angle CAB=\angle CBD$ (Since angles opposite to equal sides are equal)

$\therefore \angle CBD={55}^{\circ }$

In $△ABC$,

$\angle CBA+\angle CAB+\angle ACB={180}^{\circ }$

But, $\angle CAB=\angle CBD={55}^{\circ }$

$\Rightarrow \angle ACB={180}^{\circ }-{55}^{\circ }-{55}^{\circ }$

i.e. $\angle ACB={70}^{\circ }$

Now, in $△ACD$ and $△BCD$,

AC = BC (given)

CD = CD (common)

AD = BD ($\because$ CD bisects AB)

$\therefore △ACD\cong △BCD$ (By SSS congruence rule)

$\Rightarrow \angle DCA=\angle DCB$ (c.p.c.t.)

$\angle DCB=\frac{\angle ACB}{2}=\frac{{70}^{\circ }}{2}$

$\therefore \angle DCB={35}^{\circ }$