Question

ABC is a triangle with AB = $13$ cm, BC =$14$ cm and CA=$15$ cm. AD and BE are the altitudes from A to B to BC and AC respectively. H is the point of intersection of the AD and BE. Then the ratio of $\frac{HD}{HB}=$

• A. $\frac{3}{5}$
• B. $\frac{12}{13}$
• C. $\frac{4}{5}$
• D. $\frac{5}{9}$

Answer 1

The correct option is A $\frac{3}{5}$

According to the question,

Triangle $BEC$ and triangle $BDH$ are similar, because they have the same angles this means that the sides of these two triangles are in the same ratio.

So,

$\frac{HD}{BD}=\frac{CE}{BC}$

Note,However that $\frac{CE}{BC}$=$cosC$,

Hence$\frac{HD}{BD}$=$cosC$, so we proceed to find $cosC$ using the cosine rule,

$c^2=a^2+b^2-2ab\cos C$

${13}^2={14}^2+{15}^2-2\left(14\right)\left(15\right)cosC$

$\cos C=\frac{{13}^2-{14}^2-{15}^2}{-2\times 14\times 15}=\frac{3}{5}$

$so\frac{HD}{HB}=\frac{3}{5}$