Question

in quadrilateral ABCD side BC parallel to side AD.Diagonals AC and BD intersect each other at P. If AP=1/3 AC then prove that DP=1/2 BP

Answer 1

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,

$\angle PAD\cong \angle PCB$

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.

Answer 2

Step-by-step explanation:

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,

And,

By AA similarity postulate,

By the property of similar triangles,

Hence proved.