Question

P is a point in the interior OF triangleABC AP BP and CP meet the side BC , AC and AB in D, E and F respectively Show that PD/AD+PE/BE+DF/CF=1​

Answer 1

Answer:

If p is centroid of triangle ABC then AP:PD=2:1

AP/AD=2/3

AP/AD+BP/BD+CP/CF=2/3)×3=2

PD/AD=1/3

PD/AD+PB/BE+CP/CF=1/3)×3

=1

Step-by-step explanation:

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Answer 2

Answer:

PD / AD + PE / BE + PF / CF = 1……………(1)

Now, AP / AD = (AD - PD) / AD = {1 - (PD / AD)}.

Similarly,

BP / BE = (BE - PE) / BE = {1 - (PE / BE)}.

and

CP / CF = (CF - PF) / CF = {1 - (PF / CF)}.

Therefore, (AP / AD) + (BP / BE) + (CP / CF)

= 3 - {(PD / AD) + (PE / BE) + (PF / CF)}

= 3 - 1

= 2."

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