In a quadrilateral ABCD ,which is not a trapezium.It is known that <DAB=<ABC=60 DEGREE.moreover , <CAB=<CBD.Then A) AB=BC+CD B) AB=AD+CD C) AB=BC+AD D) AB=AC+AD ? Choose correct option.

Given

• #"In quadrilateral "ABCD, /_DAB=/_ABC=60^@#

• #/_CAB=/_CBD#

  Construction

  • # BC and AD " are produced to intersect at E"#

  • # In Delta ABE,/_EBA=/_EAB=60^@ =>/_BEA=60^@#

  • #:. Delta ABE " is equilateral"->AB=BE=EA#

    Now

    • #In DeltaABC and Delta BED, /_BED=/_CBA=60^@#
      #/_EBD=/_CAB (given)and BE= AB (proved)#
    • # DeltaABC and Delta BED " are congruent by ASA Rule"#
    • #:.DE = BC " corresponding sides"#
      #"of congruent triangles"#
    • # "Finally", AB=AE=AD+DE=AD+BC#

    So option (c) is correct

To ease the comprehension of the problem refer to the figure below:

If the sides BC and AD are extended the #triangle_(ABF)#, equilateral, is formed.

First, in addition to the angles informed it was possible to determine other interior angles of the triangles ABC and ABD, as shown in the figure.

Then using the Law of Sines on these triangles:

• #triangle_(ABC)#

#"AB"/sin (120^@-alpha)="BC"/sin alpha="AC"/sin 60^@#

• #triangle_(ABD)#

#"AB"/sin (60^@+alpha)="AD"/sin (60^@-alpha)="BD"/sin 60^@#

To focus on viable hypotheses I suggest to begin trying to prove a specific case.
For instance #AB =5# and #alpha = 15^@#:

From #triangle_(ABC) -> 5/(sin 105^@)="BC"/sin 15^@# => # BC = 5*sin 15^@/sin 105^@#

From #triangle_(ABD) -> 5/(sin 75^@)="AD"/sin 45^@# => # AD=5*sin 45^@/sin 75^@#

Note that #sin 75^@=sin 105# because #75^@# and #105^@# are supplementary angles and sines of angles of the first and second quadrants are positive.

Finally,

#AD+BC=5*(sin 15^@+sin 45^@)/(sin 75^@)=5*1=AB#

So in this case hypothesis (C) is true (I tried the others and verified that they are false.)

Now for the general proof of the hypothesis (C), following the same steps that worked for the specific proof:

From #triangle_(ABC) -> "AB"/sin (120^@-alpha)="BC"/sin alpha#

Observation:

#sin (120^@-alpha)=sin (180^@-(120^@-alpha))=sin(60^@+alpha)#

#=>"AB"/sin (60^@+alpha)="BC"/sin alpha => "BC"="AB"*sin alpha/sin (60^@+alpha)#

Therefore,

# "BC" ="AB" * (sin alpha)/((sqrt(3)/2)*cos alpha+(1/2)*sin alpha)#

From #triangle_(ABD) -> "AB"/sin (60^@+alpha)="AD"/sin(60^@-alpha)#

This will get you

#AD="AB"*sin(60^@-alpha)/sin (60^@+alpha)#

#=AB((sqrt(3)/2)*cos alpha-(1/2)*sin alpha)/((sqrt(3)/2)*cos alpha+(1/2)*sin alpha)#

Finally,

#BC+AD="AB"(sin alpha+(sqrt(3)/2)*cos alpha-(1/2)*sin alpha)/((sqrt(3)/2)*cos alpha+(1/2)*sin alpha)#

#BC + AD ="AB" * 1#

Therefore,

#AB=BC+AD# and the hypothesis (C) is true.**