Question

In the adjoining figure

$AB=AC,CH=CB$

$HK\parallel BC$

If $\angle CAD={137}^{\circ }$ , find $\angle CHK$

Answer 1

Given : $AB=AC,CH=CB$ and $HK\parallel BC$

TO FIND: $\angle CHK$

SINCE $AB=AC$

THUS, $\Delta ABC$ IS ISOSCELES.

$\angle DAC$ IS THE EXTERNAL ANGLE, WHICH EQUALS SUM OF OPPOSITE INTERIOR ANGLES:

$\angle DAC=\angle ABC+\angle ACB$

$\Rightarrow {137}^0=2\angle ABC$

$\left[\angle ABC=\angle ACB\right]$

$\Rightarrow \angle ABC={68.5}^0$

SINCE $HK\parallel BC,\angle KHB+\angle HBC={180}^0$

WHICH IMPLIES $\angle KHB={180}^0-{68.5}^0={111.5}^0$

SINCE $CH=CB$, THUS $\Delta CHB$ IS ISOSCLELES AND THUS,

$\angle CHB=\angle ABC={68.5}^0$

SINCE THE ANGLES ARE ADJACENT, $\angle CHB+\angle CHK=\angle BHK$

$\angle CHK={111.5}^0-{68.5}^0={43}^0$

$\angle CHK={43}^0$