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AP=BP,
AQ=CQ,
BD=DC,
ar(∆APQ) =ar(∆PBD) =ar(∆QDC) =ar(∆PQR) = x,
ar(∆ABC) = 4x,
from mid-point theorm,
we can say that,
given, D is the mid-point of BC, then D is passing a line parallel to AC, to AB so, P is also the mid-point of AB, similarly Q is the mid-point of AC.
Then a line passing through points P,Q is parallel to BC.
Therefore, we can conclude that, PQ // BC .
hence proved.
AQ=CQ,
BD=DC,
ar(∆APQ) =ar(∆PBD) =ar(∆QDC) =ar(∆PQR) = x,
ar(∆ABC) = 4x,
from mid-point theorm,
we can say that,
given, D is the mid-point of BC, then D is passing a line parallel to AC, to AB so, P is also the mid-point of AB, similarly Q is the mid-point of AC.
Then a line passing through points P,Q is parallel to BC.
Therefore, we can conclude that, PQ // BC .
hence proved.
Anonymous:
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it is not that much correct because I got 1mark out of 2mark for your answer
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