Question

In △ABC, M∈ AC and N∈ BC . Prove that MN ∥ AB , if: CM/MA= CN/NB

Answer 1

Answer:

See explanation

Step-by-step explanation:

In triangle △ABC, M∈ AC and N∈ BC and

$\dfrac{CM}{MA}=\dfrac{CN}{NB}=k,$

then

$CM=kMA\Rightarrow MA=\dfrac{1}{k}CM\\ \\CN=kNB\Rightarrow NB=\dfrac{1}{k}CN$

and

$AC=CM+MA=CM+\dfrac{1}{k}CM=\dfrac{k+1}{k}CM\\ \\BC=CN+NB=CN+\dfrac{1}{k}CN=\dfrac{k+1}{k}CN$

Consider triangles ABC and MNC. In these triangles,

• $AC=\dfrac{k+1}{k}CM;$
• $BC=\dfrac{k+1}{k}CN;$
• $\angle ACB\cong \angle MCN$ by reflexive property.

Hence, $\triangle ABC\sim \triangle MNC$ by SAS similarity theorem. Similar triangle have congruent corresponding angles, so

$\angle BAC\cong \angle NMC\\ \\\angle ABC\cong \angle MNC$

Angles ABC and MNC are corresponding angles when lines AB and MN are cut by transversal BC. Since these angles are congruent, by converse of  corresponding angles theorem, lines AB and MN are parallel.