Question

32. In figure, D, E and F are the three points on BC such that BD = DE = EF = FC. Show that
ar(AABD) + ar(AAEC) = 3 ar(AADE)
B
D E
C
hthat AD - RE - CE Prove that AABC is an​

Answer 1

Area of ΔABD  + Area of ΔAEC  = 3 Area of ΔADE

Step-by-step explanation:

Draw AM ⊥ BC  ( D , E & F are points on BC)

Hence AM ⊥ BD , DE, EF & FC

AM = H

BD = DE = EF = FC = B

Area of ΔABD = (1/2) BD * AM  = (1/2) BH

Area of ΔAEC = (1/2) BD * AM  = (1/2) EC * AM

= (1/2)( EF + FC) * H

= (1/2)(B + B) H

= BH

Area of ΔABD  + Area of ΔAEC  = (1/2)BH + BH  = 3BH/2

Area of ΔADE = (1/2)DE * AM  = (1/2)BH

3 Area of ΔADE = 3 * (1/2)BH = 3BH/2

3BH/2 = 3BH/2

Area of ΔABD  + Area of ΔAEC  = 3 Area of ΔADE

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