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While reviewing my maths notebooks, I came across with few unsolved problems. Here is the one of them.

ABCD is quadrilateral, such that $AD = 2$, and $\angle ABD = \angle ACD = 90^\circ $.

Point E is intersection of bisectors of triangle ABD, and point F is the intersection of bisectors of triangle ACD. $EF = \sqrt{2}$.

Find BC.

Some obvious facts:
1. ABCD is cyclic. (Let point O be midpoint of AD)
2. İntersection of BE and CF is point H. OH is perpendicular to the AD.

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Very interesting. enter image description here

Let $M$ be the midpoint of the arc $AD$ in the circumcircle of $ABCD$. Then the bisectors of $\widehat{ABD}$ and $\widehat{ACD}$ meet on $M$. Since both $\widehat{AED}$ and $\widehat{AFD}$ equals $135^\circ$, both $E$ and $F$ belong to the circle having centre $M$ through $A$ and $B$, whose radius is $\sqrt{2}$. Since $EF=\sqrt{2}$ too, we have $\widehat{BMC}=\widehat{EMF}=60^\circ$, so the angle between $B$ and $C$ in the circumcircle of $ABCD$ is $120^\circ$, and:

$$ BC = \color{red}{\sqrt{3}}.$$

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Slightly different from @Jack's approach, but with the same key goal of showing that $\overline{EF}$ is the side of an equilateral triangle. (Figure not-quite-to-scale.)

enter image description here

As bisectors of right angles $\angle ABD$ and $\angle ACD$, extended segments $\overline{BE}$ and $\overline{CF}$ meet at the midpoint $M$ of semicircle $\stackrel{\frown}{AD}$. Angles subtending arcs $\stackrel{\frown}{AM}$ and $\stackrel{\frown}{DM}$ have measure $45^\circ$. By the Exterior Angle Theorem applied to $\angle E$ of $\triangle ABE$, $$\angle AEM = \angle EAB+\angle EBA = \angle EAD + \angle DAM = \angle EAM$$ Thus, $\triangle AEM$ and (likewise) $\triangle DFM$ are isosceles with $$|\overline{EM}|=|\overline{AM}|=\sqrt{2}=|\overline{DM}|=|\overline{FM}|$$ As $|\overline{EF}|$ is given to be $\sqrt{2}$, as well, $\triangle EFM$ is equilateral, and so that $\overline{BC}$ is subtended by a $60^\circ$ angle at $M$, and a $120^\circ$ angle at $O$; thus, $\overline{BC}$ is the edge of an equilateral triangle (not shown) inscribed in the circle, and it has length $\sqrt{3}$. $\square$

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