Question

ABCD is a rhombus and AED is an equilateral triangle. E and C lie on the opposite sides of AD. If $\mathrm{\angle }ABC={78}^{\circ }$, calculate $\mathrm{\angle }DCE$and $\mathrm{\angle }ACE$.

(a) ${20}^{\circ },{30}^{\circ }$
(b) $${21}^{\circ },{31}^{\circ }$$
(c) $${22}^{\circ },{32}^{\circ }$$
(d) $${19}^{\circ },{29}^{\circ }$$

Answer 1

Hint: The key to solve the problem is to know the properties of the rhombus that opposite angles of rhombus are equal and diagonal of rhombus bisect the angles. Also, for this sum we need to know that all the angles of the equilateral triangle is ${180}^{\circ }$.

Complete step by step solution:
It is given that ABCD is a rhombus and AED is an equilateral triangle.
The angles of all equilateral triangles are $${60}^{\circ }$$.
Therefore, $$\mathrm{\angle }ADE=\mathrm{\angle }AED=\mathrm{\angle }EAD={60}^{\circ }....................(i)$$
Opposite angles of a rhombus are equal.
Therefore, $$\mathrm{\angle }ABC=\mathrm{\angle }ADC={78}^{\circ }....................(ii)$$
Also, all the sides of a rhombus are equal. By using this property ABC is an isosceles triangle with $AB=BC$.So the angles made by them will also be equal.
Therefore, $\mathrm{\angle }BAC=\mathrm{\angle }BCA...............(iii)$
By the property of the triangle that the sum of all the angles is $${180}^{\circ }$$.
In ABC,
 $$\mathrm{\angle }ABC+\mathrm{\angle }BAC+\mathrm{\angle }BCA={180}^{\circ }$$
From equation (ii) and (iii), we get,
$$\begin{array}{rl}& {78}^{\circ }+\mathrm{\angle }BAC+\mathrm{\angle }BCA={180}^{\circ }\\ & 2\mathrm{\angle }BAC={102}^{\circ }\\ & \mathrm{\angle }BAC={51}^{\circ }...............(iv)\end{array}$$
By the property that the diagonal of a rhombus is the angle bisector i.e. it divides the angle into two equal parts.
By using the above property, $\mathrm{\angle }BCA=\mathrm{\angle }ACD={51}^{\circ }................(v)$
From (i) and (ii), we know that $\mathrm{\angle }EDC=\mathrm{\angle }EDA+\mathrm{\angle }ADC$
Therefore, $\mathrm{\angle }EDC={78}^{\circ }+{60}^{\circ }={138}^{\circ }.......(vi)$
EDC is an isosceles triangle with $ED=DC$.So the angles made by them will also be equal.
Therefore, $\mathrm{\angle }DEC=\mathrm{\angle }DCE...............(vii)$
By the property of the triangle that the sum of all the angles is $${180}^{\circ }$$.
In EDC,
 $$\mathrm{\angle }EDC+\mathrm{\angle }DEC+\mathrm{\angle }DCE={180}^{\circ }$$
From equation (vi) and (vii), we get,
$$\begin{array}{rl}& {138}^{\circ }+\mathrm{\angle }DCE+\mathrm{\angle }DCE={180}^{\circ }\\ & 2\mathrm{\angle }DCE={42}^{\circ }\\ & \mathrm{\angle }DCE={21}^{\circ }...............(viii)\end{array}$$
From (v) and (viii), $$\mathrm{\angle }ACD={51}^{\circ }$$and $\mathrm{\angle }DCE={21}^{\circ }$
$\begin{array}{rl}& \mathrm{\angle }ACD=\mathrm{\angle }ACE+\mathrm{\angle }DCE\\ & {51}^{\circ }=\mathrm{\angle }ACE+{21}^{\circ }\\ & \mathrm{\angle }ACE={51}^{\circ }-{21}^{\circ }={20}^{\circ }..........................(ix)\end{array}$
Therefore, from (viii) and (ix) we get
$\mathrm{\angle }DCE={21}^{\circ },\mathrm{\angle }ACE={20}^{\circ }$ is the required solution.

Note: There is also an alternative way to find $\mathrm{\angle }ACD$. We can also find by using the properties that the adjacent angles of rhombus sum up to ${180}^{\circ }$.
Therefore, $\mathrm{\angle }ABC+\mathrm{\angle }BCD={180}^{\circ }$
$\mathrm{\angle }BCD={180}^{\circ }-{78}^{\circ }={102}^{\circ }$
By the property that the diagonal of a rhombus is the angle bisector i.e. it divides the angle into two equal parts.
By using the above property, $\mathrm{\angle }ACD={\displaystyle \frac{\mathrm{\angle }BCD}{2}}={\displaystyle \frac{{102}^{\circ }}{2}}={51}^{\circ }$.