Bunuel wrote:
imhimanshu wrote:
Bunuel wrote:
Three sides are equal) --> \(\angle{BCE}=\angle{DCE}\). So if we continue the line segment CE it will meet with the crosspoint of the diagonals, let's call it point O. Also note that EO will be perpendicular to BD, as diagonals in square make a right angle.
Hello Bunuel
Could you please explain the underlying logic behind highlighted portion. I failed to understand why it would meet at point O. Please explain.
Thanks
Since triangles BCE and CDE are congruent, then \(\angle{BEC}=\angle{DEC}\), which means that CE is the bisector of angle E. But triangle BED is an isosceles, thus bisector of angle E must also be the height and the median.
Thank you
Bunuel for providing this clear explanation!
I have practiced this question twice, and even though every time I strongly felt that extension of the line segment CE could coincide with one diagonal of the square, I could not really find a proof. I was aware that the triangles BCE and DCE are congruent, but I completely ignored that the bigger triangle BDE is an isosceles triangle, in which whatever bisects the vertex could be the height and bisect the base, and vice versa. Because the line segment CE bisects the angle E, the extension of CE, OE, could be the height of the isosceles triangle and the height bisects the base BD. Hence, we get that the length of the segment BO is half the length of BD and thus we will get the area.
By the way, since the square's diagonal AC is perpendicular to another diagonal BD and bisects BD, we can know that the segment OC is doing the same thing as does the diagonal AC. But we do not necessarily need to know this to solve this problem.
Thank you!