Question

in triangle ABC, D is the midpoint of AB and P is any point on BC if PQ parallel PD meets AB and Q in the given figure then prove that area of triangle BPQ is equal to half area of triangle ABC.​

Answer 1

$ar(BPQ) = \frac{1}{2} ar(ABC)$

Step-by-step explanation:

given : in triangle ABC , D is the mid point of of AB

proof:

CD is the median of triangle ABC

therefore ,

$ar(BCD) = ar(DAC)$

now, $\bigtriangleup PDC ,\bigtriangleup PDQ$ are on the same base PD and between the same parallel lines PD and CQ

ar(PDC) = ar(PDQ)

we know that the median divides the triangle into two triangles of equal area

therefore

$ar(BDC) = \frac{1}{2} ar(ABC)\\\\ar(BPD) +ar(PDC) =\frac{1}{2} ar(ABC)\\\\ar(BPD) +ar(PDQ) = \frac{1}{2} ar(ABC)\\\\\bigtriangleup BPD +\bigtriangleup PDQ=\bigtriangleup BPQ \\\\ar(BPQ) = \frac{1}{2}ar(ABC)$

hence proved

#Learn more:

In ABC , D is the mid - point of AB and P is any point on BC . if CQ || PD meets AB in Q in the given figure , prove that ar ( ∆ BPQ ) = at ( ∆ ABC )

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