Answer :
AB parallel to CD. So B+C = 180 degrees
AE = AD => in ΔAED, angle AED = angle EDA = let us say Ф
So angle BED = 180 - Ф angle EDC = ?
In quadrilateral BEDC, sum all angles = 360. angle B +C = 180 . So angles BED + EDC = 180.
So angle EDC = 180 - (180 - Ф) = Ф
So DE bisects ADC.
AE = AD => in ΔAED, angle AED = angle EDA = let us say Ф
So angle BED = 180 - Ф angle EDC = ?
In quadrilateral BEDC, sum all angles = 360. angle B +C = 180 . So angles BED + EDC = 180.
So angle EDC = 180 - (180 - Ф) = Ф
So DE bisects ADC.
AB II to CD.
So B+C = 180 (Linear Pair)
AE = AD (given) I'm not really sure
There fore, AED = angle EDA = let us say x
So angle BED = 180 - x angle EDC = ?
In quadrilateral BEDC, sum all angles = 360. angle B +C = 180 . So angles BED + EDC = 180. hope it helps you out
So angle EDC = 180 - (180 - x) = x
So DE bisects ADC. Yay!! Proven :P
So B+C = 180 (Linear Pair)
AE = AD (given) I'm not really sure
There fore, AED = angle EDA = let us say x
So angle BED = 180 - x angle EDC = ?
In quadrilateral BEDC, sum all angles = 360. angle B +C = 180 . So angles BED + EDC = 180. hope it helps you out
So angle EDC = 180 - (180 - x) = x
So DE bisects ADC. Yay!! Proven :P