Question #140254
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
1
Expert's answer
2020-10-26T16:54:54-0400


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

To prove: quadrilateral DBEF is a square.

Proof:

With the lettering in the diagram above,

AD^B=90°(∆DAB is a right isosceles  at D)DA^B=DB^A(base angles of an isosceles )DA^B+DB^A+AD^B=180°(angle sum of a ∆)DA^B+DB^A=180°AD^BDA^B+DB^A=180°90°DA^B+DB^A=90°DA^B+DA^B=90°(DA^B=DB^A)2DA^B=90°DA^B=45°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,

DB^C=AD^B=90°BC^D=BD^C=45°D\widehat{B}C=A\widehat{D}B=90°\\ B\widehat{C}D=B\widehat{D}C=45°\\

Since line DE bisects ADB,

BD^E=45°ED^F=BD^C+BD^E=45°+45°=90°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 DB^C,D\widehat{B}C,

DB^F=45°EB^F=DB^A+DB^F=45°+45°=90°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 (ED^F and EB^F)(E\widehat{D}F \textsf{ and } E\widehat{B}F) are right angles. DBEF is a square.


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Comments

Assignment Expert
26.10.20, 18:11

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Ali Atwi
25.10.20, 10:30

You're so late I have to send my homework in 10 minutes

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