Math, asked by nidhi291, 11 months ago

In quadrilateral ABCD, side BC parallel side AD. Diagonals AC and BD intersect each other at P. If AP = 1 upon 3 AC, then prove that DP = 1 upon 2 BP. ​


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Answers

Answered by ariestheracer
66
\bf {Given:}

In quadrilateral ABCD,

AD ║ BC and AP=1/3 AC

\bf {To\: prove: } DP=1/2 BP

\bf {Proof:}

Since, AP=\frac{1}{3} AC

\frac{AP}{AC} = \frac {1}{3}

Let, AP = x and AC = 3x

Where x is any number,

⇒ CP = AC - AP = 3x - x = 2x

Now, Inquadrilateral ABCD,

AD ║ BC

By the Alternative interior angle theorem,

\angle PAD\cong \angle PCB

And, \angle PDA\cong \angle PBC

By AA similarity postulate,

\triangle APD\sim \triangle CPB

By the property of similar triangles,

\frac{AP}{PD} = \frac{CP}{PB}

\frac{x}{PD} = \frac{2x}{PB}

\frac{PB}{PD} = \frac{2}{1}

\frac{PD}{PB} = \frac{1}{2}

PD = \frac{1}{2}\times PB

Hence proved.

Q.E.D.
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