In a convex quadrilateral ABCD, AB = BC and AD = CD. Angle B is 58 degrees and angle D is 136 degrees. Find angle A in degrees.

Step-by-step solution

1. 1

   Recognise the shape from the two equal-side conditions. Two pairs of adjacent equal sides ($AB=BC$ and $AD=CD$) make $ABCD$ a kite. The reason this matters: both $B$ and $D$ are equidistant from $A$ and from $C$, so both lie on the perpendicular bisector of $AC$. That makes the diagonal $BD$ an axis of symmetry.

2. 2

   Use the symmetry to conclude $\angle A = \angle C$. Reflecting across line $BD$ swaps $A$ and $C$ while fixing $B$ and $D$, so the angle at $A$ maps onto the angle at $C$. Hence

   $\angle A = \angle C$

   This single fact is what turns a four-unknown problem into a one-unknown problem.

3. 3

   Apply the angle sum of a quadrilateral. Every quadrilateral's interior angles add to $360^\circ$:

   $\angle A + \angle B + \angle C + \angle D = 360^\circ$

4. 4

   Substitute the known angles and the equality. Writing $\angle C$ as $\angle A$:

   $2\angle A + 58^\circ + 136^\circ = 360^\circ$

   $2\angle A = 360^\circ - 194^\circ = 166^\circ$

   $\angle A = 83^\circ$

5. 5

   Cross-check by splitting along the diagonal $AC$. Triangle $ABC$ is isosceles with apex angle $58^\circ$, so its base angles are $\tfrac{180^\circ-58^\circ}{2}=61^\circ$, giving $\angle BAC = 61^\circ$. Triangle $ACD$ is isosceles with apex $136^\circ$, so $\angle DAC = \tfrac{180^\circ-136^\circ}{2}=22^\circ$. Then $\angle A = 61^\circ + 22^\circ = 83^\circ$, which matches.