Answer to Question #140254 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

Given: ∆DAB is a right isosceles triangle at D.

To prove: quadrilateral DBEF is a square.

Proof:

With the lettering in the diagram above,

$A\widehat{D}B=90°(\textsf{∆DAB is a right isosceles ∆ at D})\\ D\widehat{A}B=D\widehat{B}A(\textsf{base angles of an isosceles ∆})\\ D\widehat{A}B+D\widehat{B}A+A\widehat{D}B=180°\\(\textsf{angle sum of a ∆)}\\ D\widehat{A}B+D\widehat{B}A=180°-A\widehat{D}B\\ D\widehat{A}B+D\widehat{B}A=180°-90°\\ D\widehat{A}B+D\widehat{B}A=90°\\ D\widehat{A}B+D\widehat{A}B=90°(D\widehat{A}B=D\widehat{B}A)\\ 2D\widehat{A}B=90°\\ D\widehat{A}B=45°\\$ Since C is placed at a point so that DABC is a parallelogram, ∆BCD has to be a right isosceles ∆ which is similar to ∆DAB. Hence,

$D\widehat{B}C=A\widehat{D}B=90°\\ B\widehat{C}D=B\widehat{D}C=45°\\$

Since line DE bisects ADB,

$B\widehat{D}E=45°\\ E\widehat{D}F=B\widehat{D}C+B\widehat{D}E=45°+45°=90°$

Also, since line BF bisects $D\widehat{B}C,$

$D\widehat{B}F=45°\\ E\widehat{B}F=D\widehat{B}A+D\widehat{B}F=45°+45°=90°\\$

Since two of the angles of quadrilateral DBEF $(E\widehat{D}F \textsf{ and } E\widehat{B}F)$ are right angles. DBEF is a square.