Prove that the altitude to the base of an isosceles triangle is also the angle bisector of the angle from which it is drawn plz ! i desperate will give 89 points and brainless answer to those who can explain as well!

Ok so isosceles triangle means all equal sides and sum of interior angles of a triangle have to add to 180deg.So if we draw a triangle with three equal 60 deg angles and draw a line from the top angle straight down to the bottom line, basically dividing the triangle into two even ones. then you can say the line or bisector line from the angle makes a 90deg with the bottom line across from angle the line is drawn out of. so then that makes two even and equal triangles, then the measure of the angles will be 90deg from bisector line + 60deg from angle untouched + 30deg from bisector angle = 180 degs for sum of interior angles in both triangles now proving the altitude from the base of an isosceles triangle is also the angle bisector of that angle. 

Does this help you understand or should I draw and upload a picture of it? Just let me know if you need anything else to understand. Thank you!

You must find the surface of the block therefore 98187 dived by 5628 = 583m squared