02 Problem involving angle and median | Properties of a Triangle

$\alpha = 180^\circ - 45^\circ$

$\alpha = 135^\circ$
 

$\beta = 180^\circ - \alpha - \theta$

$\beta = 180^\circ - 135^\circ - \theta$

$\beta = 45^\circ - \theta$
 

 

From ΔDEC
$\tan 45^\circ = \dfrac{h}{y}$

$y = \dfrac{h}{\tan 45^\circ}$

$y = h$
 

From ΔAEC
$\tan \theta = \dfrac{h}{x}$

$x = \dfrac{h}{\tan \theta}$
 

From ΔBEC
$\tan \beta = \dfrac{h}{y + z}$

$z = \dfrac{h}{\tan \beta} - y$

$z = \dfrac{h}{\tan (45^\circ - \theta)} - h$

$z = \dfrac{h}{\dfrac{\tan 45^\circ - \tan \theta}{1 + \tan 45^\circ \, \tan \theta}} - h$

$z = \dfrac{h}{\dfrac{1 - \tan \theta}{1 + \tan \theta}} - h$

$z = \dfrac{h(1 + \tan \theta)}{1 - \tan \theta} - h$
 

Since CD is median of ΔABC, point D is at the midpoint of AB, thus,
$x + y = z$

$\dfrac{h}{\tan \theta} + h = \dfrac{h(1 + \tan \theta)}{1 - \tan \theta} - h$

$\dfrac{1}{\tan \theta} + 1 = \dfrac{1 + \tan \theta}{1 - \tan \theta} - 1$

$\dfrac{1}{\tan \theta} - \dfrac{1 + \tan \theta}{1 - \tan \theta} + 2 = 0$

$(1 - \tan \theta) - \tan \theta(1 + \tan \theta) + 2\tan \theta(1 - \tan \theta) = 0$

$1 - \tan \theta - \tan \theta - \tan^2 \theta + 2\tan \theta - 2\tan^2 \theta = 0$

$1 - 3\tan^2 \theta = 0$

$\tan^2 \theta = \frac{1}{3}$

$\tan \theta = \frac{1}{\sqrt{3}}$

$\theta = 30^\circ$           answer

Triangle ABC is similar to triangle CBD
$\alpha = 180^\circ - 45^\circ = 135^\circ$
 

 

By ratio and proportion of similar triangles
$\dfrac{\text{Opposite to } \alpha}{\text{Opposite to } \theta}: ~ \dfrac{2x}{y} = \dfrac{y}{x}$

$2x^2 = y^2$

$y = \sqrt{2} x$
 

Apply Sine Law to ΔCBD
$\dfrac{y}{\sin \alpha} = \dfrac{x}{\sin \theta}$

$\dfrac{\sqrt{2} x}{\sin 135^\circ} = \dfrac{x}{\sin \theta}$

$\sin \theta = \dfrac{\sin 135^\circ}{\sqrt{2}}$

$\theta = 30^\circ$           answer