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

Step-by-step solution

1. 1

   Draw the diagonal $AC$ — it is the only line that makes both side conditions usable. The two given equalities, $AB=BC$ and $AD=CD$, each involve two sides meeting at a vertex whose angle you know. Cutting the quadrilateral along $AC$ turns each condition into an isosceles triangle with $AC$ as its base, which is what makes the angles computable. The figure is a kite with axis of symmetry $BD$.

2. 2

   Use triangle $ABC$ to split off part of angle $A$. Since $AB=BC$, triangle $ABC$ is isosceles with base $AC$, so the base angles are equal: $\angle BAC=\angle BCA$. The angle sum gives

   $2\angle BAC+58^{\circ}=180^{\circ}\ \Longrightarrow\ 2\angle BAC=122^{\circ}\ \Longrightarrow\ \angle BAC=61^{\circ}$

3. 3

   Use triangle $ADC$ for the other part. Since $AD=CD$, triangle $ADC$ is isosceles with the same base $AC$, so $\angle DAC=\angle ACD$ and

   $2\angle DAC+136^{\circ}=180^{\circ}\ \Longrightarrow\ 2\angle DAC=44^{\circ}\ \Longrightarrow\ \angle DAC=22^{\circ}$

4. 4

   Add the two pieces to rebuild angle $A$. Because the quadrilateral is convex, the diagonal $AC$ lies inside it and therefore splits $\angle A$ into exactly these two adjacent angles:

   $\angle A=\angle BAC+\angle DAC=61^{\circ}+22^{\circ}=83^{\circ}$

   Convexity is what licenses the addition — in a non-convex figure the diagonal could fall outside and the parts would subtract instead.

5. 5

   Check with the quadrilateral angle sum. The same argument applied at $C$ gives $\angle C=\angle BCA+\angle ACD=61^{\circ}+22^{\circ}=83^{\circ}$, as the kite's symmetry demands. Then

   $58^{\circ}+83^{\circ}+136^{\circ}+83^{\circ}=360^{\circ}$

   which is exactly the angle sum of a quadrilateral, so $\angle A=83^{\circ}$ is consistent.