Answer to Question #142687 in Geometry for mkami

Question

Angle bisectors of angles BAC and BCA of triangle ABC intersect its altitude BH at points P and
Q respectively. It is known that point H lies on side AC, P is situated between H and Q,
BQ = 54,5, QP = 17,5, PH = 36. Find the area S of triangle ABC. As the answer,
indicate the value of S√3 .

Answer 1

$$BH=BQ+QP+PH=$$

$$=54.5+17.5+36=108$$

The right triangle $APH$

Let $AH=x, \angle PAH=\alpha.$ Then $\tan(\alpha)=\dfrac{PH}{AH}=\dfrac{36}{x}$

The right triangle $ABH$

$\tan(2\alpha)=\dfrac{BH}{AH}=\dfrac{108}{x}$

$$\tan(2\alpha)=\dfrac{2\tan(\alpha)}{1-\tan^2(\alpha)}=\dfrac{108}{x}$$

$$\dfrac{108}{x}=\dfrac{2\cdot\dfrac{36}{x}}{1-(\dfrac{36}{x})^2 }$$ $$1-(\dfrac{36}{x})^2=\dfrac{2}{3}$$

$$x=36\sqrt{3}$$

$$\alpha=30\degree$$

The right triangle $QCH$

Let $CH=y, \angle QCH=\beta.$ Then $\tan(\beta)=\dfrac{QH}{CH}=\dfrac{53.5}{y}$

The right triangle $BCH$

$\tan(2\beta)=\dfrac{BH}{CH}=\dfrac{108}{y}$

$$\tan(2\beta)=\dfrac{2\tan(\beta)}{1-\tan^2(\beta)}=\dfrac{108}{y}$$

$$\dfrac{108}{y}=\dfrac{2\cdot\dfrac{53.5}{y}}{1-(\dfrac{53.5}{y})^2 }$$ $$1-(\dfrac{53.5}{y})^2=\dfrac{107}{108}$$

$$(\dfrac{y}{53.5})^2=108$$

$$y=321\sqrt{3}$$

Then

$$AC=AH+CH=36\sqrt{3}+321\sqrt{3}=357\sqrt{3}$$

$$S_{ABC}=\dfrac{1}{2}BH\cdot AC=$$

$$=\dfrac{1}{2}(108)(357\sqrt{3})=19278\sqrt{3}$$

$$S\sqrt{3}=57834$$