Question

In the figure below, $AB = AC = CD$. If $\angle ADB = 20^\circ$, what is the value of $\angle BAD$?

• A. $40^\circ$
• B. $60^\circ$
• C. $70^\circ$
• D. $120^\circ$
• E. $140^\circ$

Hint:

When points lie on a straight line, you can freely replace an angle with respect to $BD$ by the corresponding one with respect to $BC$. Combine this with isosceles triangles and angle sums to unlock such geometry puzzles quickly.

Answer 1

The Correct Option is D

Step 1: Use $AB=AC$.
$AB=AC \Rightarrow \triangle ABC$ is isosceles with base $BC$. Hence, $\angle ABC=\angle ACB$. Also, $B,C,D$ are collinear (from the figure), so directions $BC$ and $BD$ are the same. Thus $\angle ABD=\angle ABC$.
Step 2: Use $AC=CD$ on $\triangle ACD$.
Since $C$ lies on the straight line $BD$, the exterior angle at $C$ relative to $\triangle ABC$ is \[ \angle ACD = 180^\circ-\angle ACB. \] With $AC=CD$, $\triangle ACD$ is isosceles, so $\angle CAD=\angle ADC$.
Step 3: Link with the given $\angle ADB=20^\circ$.
At $D$, line $DB$ is collinear with $DC$, hence $\angle ADB=\angle ADC=20^\circ$. Therefore in $\triangle ACD$, \[ \angle CAD=\angle ADC=20^\circ. \] Hence \[ \angle ACD=180^\circ-\angle ACB=180^\circ- \angle ABC=180^\circ- \angle ABD. \] Step 4: Find the angles in $\triangle ABC$ and $\triangle ABD$.
In $\triangle ACD$, angle sum gives \[ \angle CAD+\angle ADC+\angle ACD=20^\circ+20^\circ+\angle ACD=180^\circ \Rightarrow \angle ACD=140^\circ. \] So $\angle ACB=180^\circ-\angle ACD=40^\circ$, and by isosceles property $\angle ABC=40^\circ$. Since $BD$ is the same direction as $BC$, $\angle ABD=\angle ABC=40^\circ$.
Step 5: Compute $\angle BAD$.
In $\triangle ABD$, \[ \angle BAD = 180^\circ - \angle ABD - \angle ADB = 180^\circ - 40^\circ - 20^\circ = 120^\circ. \] \[ \boxed{\angle BAD = 120^\circ} \]