1. Show that the area of a rhombus on hypotenuse of a right angled triangle, with one of the angles as 60 is equal to the sum of areas of rhombuses with one of their angles as 60 drawn on others sides.
2. BO and CO bisect angle B and angle C of triangle ABC. AO produced meets BC at P. Show:-
� AB*OP=BP*AO
� AC*OP=CP*AO
� AB*PC=AC*BP
� AP bisects angle BAC.
A) ΔABC is a right angle with ∠B as right ∠.
let AB = a and BC = b
By Pythagoras theorem AC = √ (a2 + b2 )
ACDE is rhombus with 60° inclination
Draw a perpendicular EF from E on base AC.
EF = AE sin 60°
AE = AC = √ (a2 + b2 )
EF = √ (a2 + b2 ) sin 60°
= √ (a2 + b2 ) /2
Area of rhombus ( ACDE ) = base * height
ACDE = AC * EF
= √ (a2 + b2 ) * √ (a2 + b2 ) √3/2
= √3/2 (a2 + b2 ) ....................( Equation 1)
similarly in rhombus BCJG
Side BC = a
angle BCJ = 60° (GIVEN)
height = a sin 60° = a √3/2
area of rhombus BCGJ = a2 √3/2 .................( equation 2)
similarly area of rhombus ABHI = b2 √3/2 ..................( equation 3)
from equation (1) (2) and (3) we have
area of rhombus ACDE = area of rhombus BCJG + area of rhombus ABHI
√3/2 (a2 + b2 ) = a2 √3/2 + b2 √3/2
√3/2 (a2 + b2 ) = √3/2 (a2 + b2 )
1 = 1
Hence proved...
B) Consider Δ APB ,
BO bisect ∠B
Therefore , AB/PB = AO/OP ( By bisect theorem..)
⇒ AB * OP = AO * PB .......................( equation 1)
Proved (1) by equation (1)
similarly AC * OP = CP * AO ..................( equation 2)
Proved (2) by equation (2)
from equation (1) and (2)
AC/CP = AB/PB
⇒ AC * PB = AB * CP ....................( equation 3)
From equation (3) we have AB/ AC = BP/PC
⇒ AB * PC = BP * AC
Therefore by bisect theorem we can say that AP bisect ∠ BAC.
Hope this helps you.
thanks.