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

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8 months 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
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8 months 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= dnow, 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)*sinxhence, area(ABC)/area(ADC)= ½ b(a+c)*sinx / ½ d(a+c)*sinx = b/d= BO/DOKINDLY APPROVE :))
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8 months ago
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