Question

in triangle DEF ,L is a point on side BC is such that LM is parallel to DFand LN is parallel to EF if MN meets ED in O when produced that prove that OL square is equal to OD into OE​

Answer 1

Proved below.

Step-by-step explanation:

Given:

In Δ DEF, side ED is produced to point O.

Also, LM║DF and LN║EF

To prove:

$OL^2=OD^2\times OE^2$

Proof:

In Δ OEM, we have, LN║EM, so

$\frac{LO}{LE}= \frac{NO}{MN}$             (By BPT theorem)

$\frac{LE}{LO} =\frac{MN}{NO}$

$\frac{LE}{LO}+1 =\frac{MN}{NO}+1$          (on adding '1' both sides)

$\frac{LE+LO}{LO} =\frac{MN+NO}{NO}$

$\frac{EO}{LO} =\frac{MO}{NO}$              (1)

In Δ OLM, we have LM║ND, so

$\frac{DO}{LD}= \frac {MN}{NO}$             (By BPT theorem)

$\frac{LD}{DO} =\frac{NO}{MN}$

$\frac{LD}{DO} +1=\frac{NO}{MN}+1$        (on adding '1' both sides)

$\frac{LD+DO}{DO} =\frac{NO+MN}{MN}$

$\frac{LO}{DO} =\frac{MO}{MN}$             (2)

So, from Eq (1) and (2), we get

$\frac{EO}{LO} =\frac{LO}{DO}$

$LO^2=OD\times OE$

Hence proved.

Answer 2

Answer:

LO²

Step-by-step explanation:

Explanation given in picture....pls use it