please answer the above question fast as I have board exam tomorrow
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ANSWER:
Given: in quadrilateral ABCD,
AD ║ BC and AP=1/3 AC
TO PROVE: DP=1/2 BP.
Proof:
Since, AP=1/3 AC
⇒ AP/AC = 1/3
Let, AP = x and AC = 3x
Where x is any number,
⇒ CP = AC - AP = 3x - x = 2x
Now, In quadrilateral ABCD,
AD ║ BC
By the Alternative interior angle theorem,
∆PAD=~∆PCB
AND ∆PDA=~∆CPB
By property of similar triangle,
AB/PD=CP/PB
X/PD=2X/PB
PB/PD=2/1
PD/PB=1/2
PD=1/2×PB
DP=1/2×BP
Hence proved
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