Would appreciate help with an explanation

In δabc shown below, ∠bac is congruent to ∠bca: triangle abc, where angles a and c are congruent given: base ∠bac and ∠acb are congruent. prove: δabc is an isosceles triangle. when completed (fill in the blanks), the following paragraph proves that line segment ab is congruent to line segment bc making δabc an isosceles triangle. (4 points) construct a perpendicular bisector from point b to line segment ac . label the point of intersection between this perpendicular bisector and line segment ac as point d: m∠bda and m∠bdc is 90° by the definition of a perpendicular bisector. ∠bda is congruent to ∠bdc by the definition of congruent angles. line segment ad is congruent to line segment dc by by the definition of a perpendicular bisector. δbad is congruent to δbcd by the line segment ab is congruent to line segment bc because consequently, δabc is isosceles by definition of an isosceles triangle. 1. corresponding parts of congruent triangles are congruent (cpctc) 2. the definition of a perpendicular bisector 1. the definition of a perpendicular bisector 2. the definition of congruent angles 1. the definition of congruent angles 2. the definition of a perpendicular bisector 1. angle-side-angle (asa) postulate 2. corresponding parts of congruent triangles are congruent (cpctc)

What is the area of the parallelogram? 48 sqrt(3)cm2 48 cm2 24 sqrt (3) cm2 24 cm2

Where would an imaginary line need to be drawn to reflect across an axis of symmetry so that a regular pentagon can carry onto itself