Proving a triangle isosceles using Thales Theorem

Question

Given that:

1. [BM) is the bisector of the angle ABC.
2. (BM) and (AN) are parallel straight lines.

I am trying to prove that the triangle ANB is an isosceles triangle with a main vertex B using Thales Theorem.

Here's what I have done:

Since BM and AN are parallel then by using Thales Theorem: $MA/MC = NB/BC = BM/AN$

I don't know where to go from here.. Any help is appreciated. Thank you.

Answer 1

Look for the alternate interior angles and corresponding angles in the figure: $\measuredangle MBA=\measuredangle BAN$ and $\measuredangle CBM=\measuredangle BNA$, so the triangle is isosceles.

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If you insist to use Thales':

$$AM/CM = BN/BC$$

From interior angle bisector theorem:

$$AM/CM=AB/BC$$

Combining the two:

$$BN/BC=AB/BC \implies AB=BN$$

Answer 2

Not using Thales's intercept theorem, but:

$AN \parallel BM$, so from $AB$ crossing $\color{#092}{\angle NAB} = \color{#02E}{\angle ABM}$ and from $CN$ crossing $\color{#092}{\angle ANB} = \color{#02E}{\angle MBC}$.

Since $BM$ bisects $\angle ABC$, $\color{#02E}{\angle ABM = \angle MBC}$ so $\color{#092}{\angle NAB =\angle ANB }$, giving $\triangle BAN$ isosceles.