Question

An altitude $BD$ and a bisector $BE$ are drawn in the triangle $ABC$ from the vertex $B$. It is known that the length of side $AC=1$ and the magnitudes of the angles $BEC,ABD,ABE,BAC$ form an arithmetic progression.

The area of circle circumscribing $\Delta ABC$ is

• A. $\frac{\pi }{8}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\pi$

Answer 1

The correct option is B $\frac{\pi }{4}$

Given $\angle BEC,\angle ABD,\angle ABE,\angle BAC$ are in Angle Property

we know

$\angle BEC=\angle ABE+\angle BAC$

(exterior angle property)

but

$\angle ABE=\angle BEC+2d\angle BAC=\angle BEC+3d\angle ABE+\angle BAC=2\angle BEC+5d\angle BEC=2\angle BEC+5d\angle BEC=-5d$

also $\angle ABb\angle BAC=90°$ (exterior angle)

$\angle BEC+d+\angle BEC+3d=90°2\angle BEC+4d=90°-\angle 0d1+4d=90°,d=-15\angle BEC=-5d-5\times -15=75°\angle ABD=60°\angle ABE=45°,\angle BAC=30°\angle ABE=45°so\angle ABC=90°$

We know $R=\frac{a}{2\sin A}=\frac{1}{2.\sin \left(\frac{\pi }{2}\right)}=\frac{1}{2}Area=\pi R^2=\frac{\pi }{4}$