Question

If in the following figure, AC = CD, AD = BD and ∠C = ${52}^o$, then the measure of ∠DAB is:

• A. ${30}^o$
• B. ${32}^o$
• C. ${34}^o$
• D. ${36}^o$

Answer 1

The correct option is B ${32}^o$

In $\Delta$ACD, AC = CD (given)
$\Rightarrow$
∠CAD = ∠CDA ( In a triangle, if two sides are equal, then the angles opposite to these sides are also equal)

Given, ∠C = ∠ACD = ${52}^o$

As the sum of all angles of a triangle is ${180}^o$, considering $\Delta$ACD, we get
∠ACD + ∠CDA + ∠CAD = ${180}^o$
52o + 2 ∠CAD = ${180}^o$
∠CAD + ∠CDA = ${64}^o$

Now, we must have,
∠CDA = ∠DAB + ∠DBA (since exterior angle is the sum of opposite interior angles)

But, ∠DAB = ∠DBA (Since AD = DB)
So, ∠DAB + ∠DAB = ${64}^o$
∠DAB = ${32}^o$