$\overline{AB}$ is a common side of two triangles $BDA$ and $BCA$. It divides the angle of the triangle $DAC$ in the ratio $1:3$. The side $\overline{DA}$ of $\Delta DAC$ is extended externally to have an exterior angle $108^\circ$.

Similarly, the length of the side $\overline{AB}$ is exactly equal to the length of the side $\overline{BD}$.

It is required to determine the value of x by using the geometric data in this geometry problem.

The line segment $\overline{AB}$ divides the $\angle DAC$ in the ratio $1:3$. Geometrically, the $\angle DAB$ is less than the $\angle BAC$.

$\dfrac{\angle DAB}{\angle BAC} = \dfrac{1}{3}$

$\implies 3 \angle DAB = \angle BAC$

$\implies \angle BAC = 3 \angle DAB$

Take $\angle DAB = y$. Therefore, $\angle BAC = 3y$.

The line segment $\overline{DE}$ is a straight line. So, it is angle is $\angle DAE$ and it is $180^\circ$ geometrically. The line segments $\overline{AB}$ and $\overline{AC}$ divided the straight angle into three sub angles. The sum of them is equal to the angle of the straight line.

$\angle DAB + \angle BAC + \angle CAE = 180^\circ$

$\implies y+3y+108^\circ = 180^\circ$

$\implies 4y = 180^\circ -108^\circ$

$\implies 4y = 72^\circ$

$\implies y = \dfrac{72^\circ}{4}$

$\therefore \,\,\,\,\,\, y = 18^\circ$

Therefore, $\angle DAB = y = 18^\circ$ and $\angle BAC = 3y = 3 \times 18^\circ = 54^\circ$

$\Delta BDA$ is a triangle. One angle $\angle DAB$ is known and it is $18^\circ$ but the other two angles $\angle ABD$ and $\angle BDA$ are unknown. It is not possible to find two angles at a time.

However, it is mentioned in our problem that the lengths of sides $\overline{AB}$ and $\overline{BD}$ are equal. In other words $AB = DB$. Therefore, the $\Delta BDA$ is an isosceles triangle. According to properties of triangle, the angles opposite to equal length sides are equal.

In $\Delta BDA$, $\overline{AB}$ and $\overline{BD}$ are sides of equal length and their opposite angles are $\angle BDA$ and $\angle DAB$ and they are equal geometrically.

$\therefore \,\,\,\,\,\, \angle BDA = \angle DAB = 18^\circ$

Now, two angles are known for the triangle $BDA$ and the third angle can be determined by applying sum of the angles of a triangle property.

$\implies \angle ABD + \angle BDA + \angle DAB = 180^\circ$

$\implies \angle ABD + 18^\circ + 18^\circ = 180^\circ$

$\implies \angle ABD + 36^\circ = 180^\circ$

$\implies \angle ABD = 180^\circ –36^\circ$

$\therefore \,\,\,\,\,\, \angle ABD = 144^\circ$

$\overline{CD}$ is a straight line and its angle $CBD$ is $180^\circ$ geometrically. The $\angle CBD$ is divided by the line segment $\angle AB$. One side of the angle $DBA$ is known and it is $144^\circ$ and the other side of the angle is unknown but it can be determined by the angle $ABD$.

$\angle ABD + \angle ABC = \angle CBD$

$\implies 144^\circ + \angle ABC = 180^\circ$

$\angle ABC = 180^\circ -144^\circ$

$\therefore \,\,\,\,\,\, ABC = 36^\circ$

The value of $x$ can be determined and it represents the angle $BCA$ in triangle $BCA$. In $\Delta BCA$, the remaining two angles are known. So, use sum of interior angles rule of a triangle to find it geometrically.

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

$\implies 36^\circ + 54^\circ + x = 180^\circ$

$\implies 90^\circ + x = 180^\circ$

$\implies x = 180^\circ -90^\circ$

$\therefore \,\,\,\,\,\, x = 90^\circ$

Latest Math Topics

Latest Math Problems

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the maths problems in different methods with understandable steps.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved