Thank you very much for the problem and its solution. An alternative solution consists of drawing a parallel to AD through C and extending BA from B to A to intersect with the parallel to AD through C, lets say at point T. Then, triangles BDA and BCT are similar, thus |BT| = 4*|BA|/3 = 4*12/3 = 16 and thus |AT| = 4. As [angle ACT] = 30° and |AC| = 2*|AT|, the angle CTA must be equal to 90° and, as a consequence, |CT| = 4*sqrt(3). This leads to [ABC] = |AB|*|CT|/2 = 12*4*sqrt(3)/2 = 24*sqrt(3).
Nice alternative solution. Thanks. Similarly, you can also draw a parallel to AC through D. And use similar triangles to show that angle BAD is 90 degrees. Rest is the same.
Thank you very much for the problem and its solution. An alternative solution consists of drawing a parallel to AD through C and extending BA from B to A to intersect with the parallel to AD through C, lets say at point T. Then, triangles BDA and BCT are similar, thus |BT| = 4*|BA|/3 = 4*12/3 = 16 and thus |AT| = 4. As [angle ACT] = 30° and |AC| = 2*|AT|, the angle CTA must be equal to 90° and, as a consequence, |CT| = 4*sqrt(3). This leads to [ABC] = |AB|*|CT|/2 = 12*4*sqrt(3)/2 = 24*sqrt(3).
Nice alternative solution. Thanks. Similarly, you can also draw a parallel to AC through D. And use similar triangles to show that angle BAD is 90 degrees. Rest is the same.