SOLUTION: This problem just confuses me to no end. Please help. M is the midpoint of AB and P is any point in AM. Line MD meeting BC at D is parallel to PC. Prove that triangle BPD has half

Algebra ->  Triangles -> SOLUTION: This problem just confuses me to no end. Please help. M is the midpoint of AB and P is any point in AM. Line MD meeting BC at D is parallel to PC. Prove that triangle BPD has half      Log On


   

Question 31174: This problem just confuses me to no end. Please help.
M is the midpoint of AB and P is any point in AM. Line MD meeting BC at D is parallel to PC. Prove that triangle BPD has half the area of triangle ABC.
This que
In triangle ABC, stion is so hard. I really hope you can find a solution! Thank you!!!
Answer by venugopalramana(3286) About Me  (Show Source):
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This problem just confuses me to no end. Please help.
In triangle ABC, M is the midpoint of AB and P is any point in AM. Line MD meeting BC at D is parallel to PC. Prove that triangle BPD has half the area of triangle ABC.
OK ...P IS ANY POINT ON AM.LET
MP=X*MA=X*AB/2..SINCE M IS MIDPOINT OF AB.
IN TRIANGLE BPC,WE HAVE MD || PC.HENCE
BD/DC=BM/MP=(AB/2)/(X*AB/2)=1/X
BD/DC=1/X.....OR....BD/(BD+DC)=1/(1+X)
BD/BC=1/(1+X)...OR.....BD=BC/(1+X)................................I
SIMILARLY FROM BM/MP = 1/X.....OR.......BM/(BM+MP)=1(1+X)
BM/BP=1/(1+X)
(AB/2)/BP=1/(1+X)......OR.....BP=AB*(1+X)/2.....................II
NOW AREA OF TRIANGLE = (1/2)*BASE*ALTITUDE
AREA OF TRIANGLE ABC = A(ABC)SAY =(1/2)BC*AB*SIN(B)..................III
SINCE ALTITUDE =AB*SIN(B).......
SIMILARLY.....
AREA OF TRIANGLE BPD=A(BPD) SAY =(1/2)BD*BP*SIN(B)........................IV
DIVIDING EQN.III/EQN.IV.....WE GET.....
A(ABC)/A(BPD)=BC*AB/BD*BP..............................................V
BUT BD=BC/(1+X) AND BP =AB*(1+X)/2...SO
BD*BP=BC*AB*(1+X)/2(1+X)=BC*AB/2...SUBSTITUTING IN EQN.V WE GET
A(ABC)/A(BPD)=BC*AB/(BC*AB/2)=2.....PROVED