Question

In triangle ABC,seg MN parallel to side AC.seg MN divides triangle ABC into two parts equal in area .determine AM/MB

Answer 1

Just change the values.You will get the answer.
Make it as brainliest!

Answer 2

Answer:

√2

Step-by-step explanation:

in ΔABC and ΔAMN

segment MN ║ BC

∠AMN=∠ABC (corresponding angle)

∠ANM=∠ACB (corresponding angle)

∠A si common

so ΔABC≅ΔAMN (CPCT)

given that 2*(area of ΔAMN )= area of ΔABC

so

$\frac{AM}{MB}=\sqrt{\frac{area of ABC}{area of AMN}}\\\\\frac{AM}{MB}=\sqrt{\frac{2}{1}}\\\\\frac{AM}{MB}=\sqrt{2}$