57. ABCD is a parallelogram. Side AB is produced on both sides to E & F as in figure such that BE = BC & AF = AD. Show that EC & FD when produced meets at right angle.
Answers
ANSWER :
WE KNOW THAT,
∠DAB + ∠ABC = 180°
[ ABCD IS A PARALLELOGRAM ]
BY EXTERIOR ANGLE PROPERTY,
∠DFA + ∠FDA + ∠CEB + ∠BCE = 180° >> (i)
[ since ∠DFA + ∠FDA = ∠DAB and ∠CEB + ∠BCE = ∠ABC]
AS BE = BC AND AF = AD,
∠CEB = ∠BCE AND ∠DFA = ∠FDA
[property of isosceles triangle]
IN (i),
2∠DFA + 2∠CEB = 180°
SINCE 180° = 720° - 540°,
2∠DFA + 2∠CEB = 720 - 540
=> 720 - 2∠DFA - 2∠CEB = 540
=> 2(360) - 2∠DFA - 2∠CEB = 360+180
=> 2(360 - ∠DFA - ∠CEB) = 360+180 >> (ii)
IN TRAPEZIUM DCEF,
IN TRAPEZIUM DCEF, ∠FDC + ∠ECD + ∠CEB + ∠DFA = 360°
IN TRAPEZIUM DCEF, ∠FDC + ∠ECD + ∠CEB + ∠DFA = 360°=> ∠FDC + ∠ECD = 360° - ∠DFA - ∠CEB >> (iii)
BY SUBSTITUTING (iii) in (ii), we get :
2(∠FDC + ∠ECD) = 360+180
=> 2∠FDC + 2∠ECD = 360+180
=> 360 - 360 + 2∠ECD - 360 + 2∠FDC = 180°
>>> (iv)
WHEN WE COMBINE - 360 + 2∠ECD, WE GET
WHEN WE COMBINE - 360 + 2∠ECD, WE GET -2(180 - ∠ECD) AND WHEN WE COMBINE - 360 + 2∠FDC, WE GET -2(180 - ∠FDC)
SO, IN (iv),
360 - 2(180 - ∠ECD) - 2(180 - ∠FDC) = 180° >> (v)
SINCE 180 - ∠ECD = ∠GCD AND 180 - ∠FDC = ∠GDC, IN (v), we get :
360 - 2∠GCD - 2∠GDC = 180°
=> 2(180° - ∠GCD - ∠GDC) = 180° >>> (vi)
SINCE ∠FGE + ∠GCD + ∠GDC = 180°,
∠FGE = 180 - ∠GCD - ∠GDC >>> (vii)
BY SUBSTITUTING (vii) in (vi), we get :
2∠FGE = 180°
=> ∠FGE = 90° >>>>>>>>>> PROVED!
Step-by-step explanation:
ANSWER :
WE KNOW THAT,
∠DAB + ∠ABC = 180°
[ ABCD IS A PARALLELOGRAM ]
BY EXTERIOR ANGLE PROPERTY,
∠DFA + ∠FDA + ∠CEB + ∠BCE = 180° >> (i)
[ since ∠DFA + ∠FDA = ∠DAB and ∠CEB + ∠BCE = ∠ABC]
AS BE = BC AND AF = AD,
∠CEB = ∠BCE AND ∠DFA = ∠FDA
[property of isosceles triangle]
IN (i),
2∠DFA + 2∠CEB = 180°
SINCE 180° = 720° - 540°,
2∠DFA + 2∠CEB = 720 - 540
=> 720 - 2∠DFA - 2∠CEB = 540
=> 2(360) - 2∠DFA - 2∠CEB = 360+180
=> 2(360 - ∠DFA - ∠CEB) = 360+180 >> (ii)
IN TRAPEZIUM DCEF,
IN TRAPEZIUM DCEF, ∠FDC + ∠ECD + ∠CEB + ∠DFA = 360°
IN TRAPEZIUM DCEF, ∠FDC + ∠ECD + ∠CEB + ∠DFA = 360°=> ∠FDC + ∠ECD = 360° - ∠DFA - ∠CEB >> (iii)
BY SUBSTITUTING (iii) in (ii), we get :
2(∠FDC + ∠ECD) = 360+180
=> 2∠FDC + 2∠ECD = 360+180
=> 360 - 360 + 2∠ECD - 360 + 2∠FDC = 180°
>>> (iv)
WHEN WE COMBINE - 360 + 2∠ECD, WE GET
WHEN WE COMBINE - 360 + 2∠ECD, WE GET -2(180 - ∠ECD) AND WHEN WE COMBINE - 360 + 2∠FDC, WE GET -2(180 - ∠FDC)
SO, IN (iv),
360 - 2(180 - ∠ECD) - 2(180 - ∠FDC) = 180° >> (v)
SINCE 180 - ∠ECD = ∠GCD AND 180 - ∠FDC = ∠GDC, IN (v), we get :
360 - 2∠GCD - 2∠GDC = 180°
=> 2(180° - ∠GCD - ∠GDC) = 180° >>> (vi)
SINCE ∠FGE + ∠GCD + ∠GDC = 180°,
∠FGE = 180 - ∠GCD - ∠GDC >>> (vii)
BY SUBSTITUTING (vii) in (vi), we get :
2∠FGE = 180°
=> ∠FGE = 90° >>>>>>>>>> PROVED!