Scale factor lab

<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>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Prove that</div><ol class="abc-list"><li data-list-item-id="item-i-a" data-list-item-path="item-i/item-i-a"><div class="list-item-paragraph">triangles \(BNF\) and \(BMD\) are similar, and</div></li><li data-list-item-id="item-i-b" data-list-item-path="item-i/item-i-b"><div class="list-item-paragraph">triangles \(BEN\) and \(BAM\) are similar.</div></li></ol></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Hence prove that</div><ol class="abc-list"><li data-list-item-id="item-ii-a" data-list-item-path="item-ii/item-ii-a"><div class="list-item-paragraph">\(N\) is the midpoint of \(EF\), and</div></li><li data-list-item-id="item-ii-b" data-list-item-path="item-ii/item-ii-b"><div class="list-item-paragraph">\(NF^2 = AN \times NC\).</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 \(BC\).

Proofs in Plane Geometry - Similarity

Proofs in Plane Geometry - Similarity

Video Solution:

Solution:

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