if the diagonals of a quadrilateral abcd meet o, then prove that abc: adc=bo:od

Arun
25757 Points
3 years ago
Given

ABCD is a parallelogram

AC and BD are diagonals

O is a mid point

To prove BO =OD

O=o common point

Angle b=angle d opp angles in ll gm are                     equal

Ob=od sides opp to equal angles are equal

Hence proved
2081 Points
3 years ago
note that aruns ans above is absurd and wrong coz he has assumed abcd to be a || gm, even though it is nowhere mentioned in the ques. we will prove the result correctly for any quad abcd as follows:
first note that area of any triangle PQR= ½ PQ*PR*sin(angle QPR) (this is a std formula nd u can find multiple proofs of this online too!)
also, let angle AOB= x, whence angle DOC= x and angle AOD= angle BOC= pi – x.
further AO= a, BO= b, CO= c and DO= d
now, area(ABC)= area(AOB) + area(BOC)= ½ ab*sinx + ½ bc*sin(pi – x)= ½ b(a+c)*sinx since sinx= sin(pi – x)
similarly area(ADC)= area(AOD) + area(COD)= ½ ad*sin(pi – x) + ½ cd*sinx= ½ d(a+c)*sinx
hence, area(ABC)/area(ADC)= ½ b(a+c)*sinx / ½ d(a+c)*sinx
= b/d
= BO/DO
KINDLY APPROVE :))