Question

In the given figure, if $AB=AC,CH=CB$ and $HK\parallel BC$, the value of $\angle HCK$ is:

• A. ${30}^o$
• B. ${35}^o$
• C. ${40}^o$
• D. ${45}^o$

Answer 1

The correct option is A ${30}^o$

Extend $HA$ to $E$. Given, $\angle EAC={140}^{\circ }$

$EAH$ is a straight line.

$\therefore \angle EAK+\angle KAH={180}^{\circ }$

$\angle KAH={180}^{\circ }-{140}^{\circ }={40}^{\circ }$

$AB=AC⟹\angle ABC=\angle ACB$

In $△ABC$,

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

$2\angle CBA={180}^{\circ }-{40}^{\circ }$

$\angle CBA=\angle ACB={70}^{\circ }$ ---(i)

$CB=CH⟹\angle CBH=\angle BHC$

In $△BHC$,

$\angle BHC+\angle CBH+\angle HCB={180}^{\circ }$

$\angle HCB={180}^{\circ }-{70}^{\circ }-{70}^{\circ }$ ---from (i)

$\angle HCB={40}^{\circ }$

$\angle BCK=\angle BCH+\angle HCK$

$\angle HCK=\angle BCK-\angle HCB$

$\angle HCK={70}^{\circ }-{40}^{\circ }={30}^{\circ }$