Question

In a triangle ABC, AD is perpendicular to BC and DE is perpendicular to AB

Answer 1

In $△BDE\angle BDE=90-\angle B$
And in $△AED\angle ADE=\angle B$ as $\angle ADE+\angle BDE=90$
A) In $△ABDAD=C\sin B$ and $BD=C\cos B$ ...(1)
Therefore area of $△ABD=\frac{1}{2}AD\times BD=\frac{1}{2}{c}^2\sin B\cos B=\frac{1}{4}{c}^2\sin 2B$
B) In $△ACDAD=B\sin C$ and $CD=b\cos C$
Therefore area of $△ACD=\frac{1}{2}\times AD\times CD=\frac{1}{2}{b}^2\sin C\cos C=\frac{1}{4}{b}^2\sin 2C$
C) In $△ADEAE=AD\sin \left(ADE\right)=AD\sin B$ and $ED=AD\cos \left(ADE\right)=AD\cos B$
Therefore area of $△ADE=\frac{1}{2}\times AE\times ED\frac{1}{2}AD\sin B\times AD\cos B$
Substituting values from (1)
$△ADE=\frac{1}{4}{c}^2{\sin }^{2}B\sin 2B$
D) In $△BDEDE=BD\sin B$ and $BE=BD\cos B$
Therefore area of $△BDE=\frac{1}{2}\times DE\times BE=\frac{1}{2}\times BD\sin B\times BD\cos B$
Substituting value from (1)
$△BDE=\frac{1}{4}{C}^2{\cos }^{2}B\sin 2B$