Question

i will mark as brainliest and i will give thanks please help

Answer 1

See attachment

Prove that figure ABCD is similar to figure AEFD

∠ABC = ∠BEF (Corresponding angles)

∠BCD = ∠EFD (Corresponding angles)

∠BAD = ∠BAD (Common angles)

∠ADC = ∠ADC (Common angles)

Therefore figure ABCD is similar to figure AEFD

By formula:

$\dfrac{\text {Length A1}}{\text{Length B1}} = \dfrac{\text {Length A2}}{\text{Length B2}}$

Therefore:

$\dfrac{\text {EF}}{\text{BC}} = \dfrac{\text {NO}}{\text{MN}}$

$\dfrac{\text {EF}}{\text{7}} = \dfrac{\text {1.5}}{\text{3}}$

$3 \times EF = 1.5 \times 7$

$3 \times EF = 10.5$

$EF = 10.5 \div 3$

$EF = 3.5$

Answer: EF = 3.5

Answer 2

Refer to the attachments!