Question

in trapezium abcd, side bc is parallel to side ad, diagonal ac and bd, intersect each other at point p .if ap is 1/3ac,then show that dp=1/3bp​

Answer 1

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