Question

In A ABC, seg MN | side AC. Seg MN divide triangle ABC into two parts equal in area. Determine AM/MB.​

Answer 1

Answer:

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

\begin{lgathered}\frac{AM}{MB}=\sqrt{\frac{area of ABC}{area of AMN}}\\\\\frac{AM}{MB}=\sqrt{\frac{2}{1}}\\\\\frac{AM}{MB}=\sqrt{2}\end{lgathered}

MB

AM

=

areaofAMN

areaofABC

MB

AM

=

1

2

MB

AM

=

2

Answer 2

Step-by-step explanation:

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