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In triangle BEC,
angle EBC + angle BCE = angle BED. ...(I)
(since sum of two opposite internal angles of a triangle equal to the exterior of the third angle)
since AD = BE is given
and ABCD is a trapezium,
ABED is an isosceles trapezium.
angle ADB = angle BED. ...(II)
Substituting equation (I) with (II),
angle EBC + angle BCE = angle ADE
(since angle ADB = angle ADE)
HENCE PROVED!
angle EBC + angle BCE = angle BED. ...(I)
(since sum of two opposite internal angles of a triangle equal to the exterior of the third angle)
since AD = BE is given
and ABCD is a trapezium,
ABED is an isosceles trapezium.
angle ADB = angle BED. ...(II)
Substituting equation (I) with (II),
angle EBC + angle BCE = angle ADE
(since angle ADB = angle ADE)
HENCE PROVED!
Answered by
0
hey it can be easily solved
see here carefully as given AD=BE
so the trapezium ABCD becomes ABED and it is isosceles trapezium because its non parallel sides are equal means angle ADE = angle BDE
....(I)
AND we are known that angle BED = angle BCE + angle EBC ( from exterior angle property)
since angle ADE = angle BDE from (I)
now , angle ADE = angle BCE + angle EBC
Hence proved.
see here carefully as given AD=BE
so the trapezium ABCD becomes ABED and it is isosceles trapezium because its non parallel sides are equal means angle ADE = angle BDE
....(I)
AND we are known that angle BED = angle BCE + angle EBC ( from exterior angle property)
since angle ADE = angle BDE from (I)
now , angle ADE = angle BCE + angle EBC
Hence proved.
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