Question

in the figure the lines AP and BR are parallel. The distance between them is 3 cm and BC = EF= QR = 4 cm​

Answer 1

Answer:

768teyyee5td65eteueye8eye

Answer 2

Answer:

Given: In ∆ABC, P is the mid point of BC. PQ||CA, PQ meets AB in Q. QR||BC, QR meets AP in R.

To prove: 1. AP = 2 AR

2. BC = 4 QR

Proof:

In ∆ABC, P is the mid point of BC and PQ||AB.

∴ Q is the mid point of AB (Converse of mid-point theorem)

In ∴ ABP, Q is the mid point of AB and QR||BP.

∴ R is the mid point of AP. (Converse of mid point theorem)

⇒ AP = 2AR