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.

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