Question

ABCD is a parallelogram of which no angle is 60º. Equilateral triangles ADE and DCF are
drawn outwardly on the sides AD and DC. Show that triangleABE is congruent to triangleCFB.​

Answer 1

Answer:

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answer with explanation...

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Answer 2

Concept

A parallelogram is a straightforward quadrilateral with two sets of parallel sides in Euclidean geometry. In a parallelogram, the opposing or confronting sides are of equal length, and the opposing angles are of equal size.

Given

ABCD is a parallelogram of which no angle is 60º. Equilateral triangles ADE and DCF are drawn outwardly on the sides AD and DC is given

Find

We have to show that ΔABE is congruent to ΔCFB

Solution

The steps are as follow:

• ABCD is a parallelogram so AB  = CD and AD = BC because in parallelogram opposite sides are equal
• Equilateral triangles have been drawn on AD and DC, respectively, as ΔADE and ΔDCF.
• Considering the ∆ ABE and ∆ CFB

AB = CD     (Opposite sides of parallelogram are equal)

AB= CF       (Sides of an equilateral triangle)

• So

∠EAB = ∠EAD + ∠DAB

∠EAB =  60º + ∠DAB             ( ∵ ∆ADE is an equilateral triangle)

∠EAB  = 60º + ∠DCB            (Opposite  angles of parallelogram are equal)

∠EAB   = ∠FCB

AE = AD    (Sides of an equilateral triangle are equal )

AE = BC     (Opposite sides of parallelogram are equal)

• In ∆ABE and  ∆CFB

AB= CF

∠EAB   = ∠FCB

AE = BC

• So congruency between triangle ΔABE and ΔCFB is proved

Hence we have showed that ΔABE is congruent to ΔCFB

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