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,
ADB=90°(∆DAB is a right isosceles ∆ at D)DAB=DBA(base angles of an isosceles ∆)DAB+DBA+ADB=180°(angle sum of a ∆)DAB+DBA=180°−ADBDAB+DBA=180°−90°DAB+DBA=90°DAB+DAB=90°(DAB=DBA)2DAB=90°DAB=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,
DBC=ADB=90°BCD=BDC=45°
Since line DE bisects ADB,
BDE=45°EDF=BDC+BDE=45°+45°=90°
Also, since line BF bisects DBC,
DBF=45°EBF=DBA+DBF=45°+45°=90°
Since two of the angles of quadrilateral DBEF (EDF and EBF) are right angles. DBEF is a square.
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26.10.20, 18:11
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25.10.20, 10:30
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You're so late I have to send my homework in 10 minutes