Question

A road map of a village is shown below. Roads AB and BD have the same length. Roads BC and CD also have the same length. What is the measure of ∠BAD ?

• A. ${30}^{\circ }$
• B. ${10}^{\circ }$
• C. ${40}^{\circ }$
• D. ${20}^{\circ }$

Answer 1

The correct option is D ${20}^{\circ }$

Let's connect the roads to make a triangle for better understanding.
Let ∠BAD be x.

ABD is an isosceles triangle.
So, AB = BD

$\text{Hence,}\angle \text{BAD}=\angle \text{BDA}$
$\Rightarrow \angle \text{BDA}=x$

BCD is an isosceles triangle.
So, BC = DC

$\text{Hence,}\angle \text{CBD}=\angle \text{CDB}$

$\text{In}△\text{BCD},\text{by angle sum property,}$ we have:
$\angle \text{BCD}+\angle \text{CDB}+\angle \text{DBC}={180}^{\circ }$
$\Rightarrow {100}^{\circ }+\angle \text{DBC}+\angle \text{DBC}={180}^{\circ }$
$\Rightarrow {100}^{\circ }+2\angle \text{DBC}={180}^{\circ }$
$\Rightarrow 2\angle \text{DBC}={180}^{\circ }–{100}^{\circ }$
$\Rightarrow 2\angle \text{DBC}={80}^{\circ }$
$\Rightarrow \angle \text{DBC}={40}^{\circ }$

So,
$\angle \text{ABD}={180}^{\circ }–\angle \text{DBC}$
$\Rightarrow \angle \text{ABD}={180}^{\circ }–{40}^{\circ }$
$\Rightarrow \angle \text{ABD}={140}^{\circ }$

$\text{In}△\text{ABD},\text{by angle sum property,}$ we have:
$\angle \text{BAD}+\angle \text{ADB}+\angle \text{DBA}={180}^{\circ }$
$\Rightarrow x^{\circ }+x^{\circ }+{140}^{\circ }={180}^{\circ }$
$\Rightarrow 2x^{\circ }={180}^{\circ }–{140}^{\circ }$
$\Rightarrow 2x^{\circ }={40}^{\circ }$
$\Rightarrow x^{\circ }={20}^{\circ }$
$\Rightarrow \angle \text{BAD}={20}^{\circ }$