<p>In triangle \(ABC\), angle \(BAC = 90^\circ\), \(AD\) is perpendicular to \(BC\) and \(M\) is the midpoint of \(AD\). \(BMN\), \(BAE\) and \(ENF\) are straight lines where&nbsp;\(ENF\) is perpendicular to \(BC\).</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/public/question-images/pasted/1770112049226-2p13pa7la4a.png" style="max-width: min(200px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Prove that</span></div><ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">triangles \(BNF\) and \(BMD\) are similar, and</span></div></li><li><div class="list-item-content"><span class="list-item-text">triangles \(BEN\) and \(BAM\) are similar.</span></div></li></ol></li><li><div class="list-item-content"><span class="list-item-text">Hence prove that</span></div><ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">\(N\) is the midpoint of \(EF\), and</span></div></li><li><div class="list-item-content"><span class="list-item-text">\(NF^2 = AN \times NC\).</span></div></li></ol></li></ol>

In triangle \(ABC\), angle \(BAC = 90^\circ\), \(AD\) is perpendicular to \(BC\) and \(M\) is the midpoint of \(AD\). \(BMN\), \(BAE\) and \(ENF\) are straight lines where \(ENF\) is perpendicular to \(BC\).

triangles \(BNF\) and \(BMD\) are similar, and

triangles \(BEN\) and \(BAM\) are similar.

Proofs in Plane Geometry

Question 5049b266