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In triangle $ABC$, $BA=BC$ and $\angle B=90^\circ$. $D,E$ are the points on $AB,BC$ respectively such that $AD=CE$. $M,N$ are points on $AC$ such that $DM$ is perpendicular to $AE$ and $BN$ is perpendicular to $AE$. Prove that $MN=NC$.

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How can we solve this geometric problem by application of midpoint theorem?

Midpoint theorem states that in triangle ABC if D,E are midpoints of AB,AC respectively then AB is parallel to BC and AB =BC/2.

And converse of midpoint theorem states that:- In triangle ABC if D is the midpoint of AB and a line say U is drawn passing through D parallel to BC then U intersects AC(Say at E) and A-E-C and AE=EC.

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    I added a sketch that should make your question a bit easier to grasp. However, I'm still puzzled about *How can we solve this sum by application of midpoint theorem?* What *sum*? And maybe it's my lack of acquaintance with English geometry terminology, but what is the *midpoint theorem*?2012-05-23
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    SUm(meaning:- this particular question.)2012-05-23
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    Midpoint theorem states that in triangle ABC if D,E are midpoints of AB,AC respectively then AB is parallel to BC and AB =BC/2.2012-05-23
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    and converse of midpoint theorem states that:- In triangle ABC if D is the midpoint of AB and a line say U is drawn passing through D parallel to BC then U intersects AC(Say at E) and A-E-C and AE=EC2012-05-23
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    So I wanted to say that whether we can prove the required result of this geometric problem using the above two theorems2012-05-23
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    Okay, thank you for the clarification, maybe you could add that information to the body of the question?2012-05-23

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