Question

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. ​

Answer 1

$\bf {Given:}$

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.