Answer to Question #140299 in Geometry for Ali Atwi

Question

BAB is a right isosceles triangle at D

1)Place point C so that DABC would be a parallelogram

2)The bisector of angle ADB cots AB at E and the bisector of angle DBC cuts CD at F

Prove that quadrilateral DEBF is a square

Answer 1

Using a ruler to measure,

$|DE| = 4.5cm\\ |EB| = 4.5cm\\ |BF| = 4.5cm\\ |FD| = 4.5cm$

$\therefore |DE| = |EB| = |BF| = |FD| = 4.5cm$

$\therefore$ In $\square DEBF$, all sides are equal

From the diagram,

$|DE| \textsf{ is the perpendicular bisector of } |AB|\\ \therefore \angle DEB = 90°$

$\textsf{Also, } |BF| \textsf{ is the perpendicular bisector of } |DC|\\ \therefore \angle BFD= 90°$

$\angle BFD = \angle DEB = 90°$

$\therefore$ Opposite angles are right angles

$\textsf{Since all the sides in } \square DEBF \textsf{ are equal and it's opposite angles are right angles }\\ \therefore \textsf{All its sides are congruent}\\ \therefore \square DEBF \textsf{ is a square}$