10/04/2024
$D$ is a point inside $\triangle{ABC}$ such that $\angle{ABD}=6^{\circ}$, $\angle{BAD}=30^{\circ}$, $\angle{CAD}=54^{\circ}$, $\angle{ACD}=24^{\circ}$, find $\angle{CBD}$.
Solve:
\(\begin{multline}
\shoveleft \text{Extend }CD \text{ to }E \text{ such that }AC=AE \implies \angle{AEC}=\angle{ACE}=24^{\circ}, \angle{EAB}=48^{\circ}\\
\shoveleft \implies \angle{EDA}=78^{\circ}=\angle{EAD}\implies EA=ED=AC, \angle{EDB}=66^{\circ}\\
\shoveleft \text{Let }F \text{ on }AB \text{ such that }\angle{DEF}=60^{\circ}\implies \angle{AEF}=84^{\circ}\implies \angle{AFE}=48^{\circ}=\angle{EAB}\\
\shoveleft \implies EF=EA=ED\implies \triangle{DEF}\text{ is equilateral}\implies \angle{EFD}=\angle{EDF}=60^{\circ}\\
\shoveleft DF=EF=DE=EA=AC\implies \angle{AFD}=12^{\circ}\implies \angle{BDF}=6^{\circ}=\angle{ABD}\\
\shoveleft \implies BF=DF=EF\implies \angle{EBF}=\angle{BEF}=24^{\circ}=\angle{ACD}\implies ACBE\text{ is cyclic}\\
\shoveleft \implies \angle{ABC}=\angle{AEC}=24^{\circ}\implies \angle{CBD}=24^{\circ}-6^{\circ}=\bbox[5px, border: 1px solid black]{18^{\circ}}
\end{multline}\)
10/06/2024
$D$ is a point inside $\triangle{ABC}$ such that $\angle{BAD}=20^{\circ}$, $\angle{CAD}=30^{\circ}$, $\angle{ACD}=40^{\circ}$, $\angle{BCD}=10^{\circ}$, find $\angle{CBD}$.
Solve:
\(\begin{multline}
\shoveleft \angle{BAC}=50^{\circ}=\angle{BCA}\implies AB=BC, \angle{ABC}=80^{\circ}\\
\shoveleft \text{Extend }AD \text{ to }E \text{ such that }AB=BE\implies \angle{BEA}=\angle{BAE}=20^{\circ}\\
\shoveleft \implies \angle{ABE}=140^{\circ}\implies \angle{EBC}=60^{\circ}\\
\shoveleft AB=BE=BC\implies \triangle{BCE} \text{ is equilateral} \implies BE=EC=BC=AE\\
\shoveleft \angle{BEC}=60^{\circ}=\angle{BCE}\implies \angle{DEC}=40^{\circ},\angle{DCE}=70^{\circ}\\
\shoveleft \implies \angle{CDE}=70^{\circ}=\angle{DCE}\implies CE=DE=BE\implies \angle{DBE}=80^{\circ}\\
\shoveleft \implies \angle{CBD}=\angle{DBE}-\angle{EBC}=\bbox[5px, border: 1px solid black]{20^{\circ}}
\end{multline}\)
10/07/2024
$\triangle{ABC}$ is an isosceles triangle with $AB=AC$, $\angle{ABC}=\angle{ACB}=80^{\circ}$, $D$ is inside $\triangle{ABC}$ such that $\angle{ABD}=50^{\circ}$, $\angle{ACD}=25^{\circ}$ , find $\angle{CAD}$.
Solve:
10/13/2024
$D,E,F$ are on side $BC, CA, AB$ of $\triangle{ABC}$ respectively such that $\angle{AFE}=\angle{BFD}$, $\angle{AEF}=\angle{CED}$, $\angle{BDF}=\angle{CDE}$, and $[AEF]=5$,$[BDF]=2$, $[CDE]=1$, find $[DEF]$.
Solve:
\(\begin{multline}
\shoveleft \text{Let } \angle{BDF}=\angle{CDE}=\angle{1}, \angle{AFE}=\angle{BFD}=\angle{2}, \angle{AEF}=\angle{CED}=\angle{3}\\
\shoveleft \implies \angle{A}+\angle{2}+\angle{3}=180^{\circ},\angle{B}+\angle{1}+\angle{2}=180^{\circ}, \angle{C}+\angle{1}+\angle{3}=180^{\circ}\\
\shoveleft \implies 2(\angle{1}+\angle{2}+\angle{3})=360^{\circ}\implies \angle{1}+\angle{2}+\angle{3}=180^{\circ}\\
\shoveleft \implies \angle{A}=\angle{1}, \angle{B}=\angle{3}, \angle{C}=\angle{2} \implies \triangle{AFE}\sim\triangle{DFB}\sim\triangle{DCE}\sim\triangle{ACB}\\
\shoveleft \text{Let }x=\sqrt{\dfrac{[DEC]}{[ABC]}}, y=\sqrt{\dfrac{[DBF]}{[ABC]}}, z=\sqrt{\dfrac{[AEF]}{[ABC]}}, AB=c, BC=a, CA=b \implies\\
\shoveleft EF=xa, AF=xb, AE=xc, FB=ya, DF=yb, BD=yc, EC=za, CD=zb, DE=zc\\
\shoveleft \implies BD+DC=BC=yc+zb=a \implies c=\dfrac{1}{y}a-\dfrac{z}{y}b\\
\shoveleft AE+EC=AC=xc+za=b \implies (\dfrac{x}{y}+z)a = (1+\dfrac{xz}{y})b\\
\shoveleft AF+FB=AB=xb+ya=c \implies (\dfrac{1}{y}-y)a=(x+\dfrac{z}{y})b\\
\shoveleft \implies \dfrac{x+yz}{1-y^2}=\dfrac{y+xz}{xy+z} \implies x^2+y^2+z^2+2xyz=1\\
\shoveleft \text{Since }x^2+y^2+z^2+\dfrac{[DEF]}{[ABC]}=1 \implies \dfrac{[DEF]}{[ABC]}=2xyz\\
\shoveleft \text{Let }s=[ABC], s_1=[AEF], s_2=[BDF], s_3=[DEC]\\
\shoveleft \implies 2\sqrt{\dfrac{s_1s_2s_3}{s^3}}=1-\dfrac{s_1+s_2+s_3}{s}\\
\shoveleft \implies s^3-2(s_1+s_2+s_3)s^2+(s_1+s_2+s_3)s-4s_1s_2s_3=0\\
\shoveleft \text{Let }m=s_1+s_2+s_3, n=s_1s_2s_3 \implies s^3-2ms^2+m^2s-4n=0\\
\shoveleft \implies s = \dfrac{\sqrt[3]{6 \sqrt{81 n^2 - 3m^3n} - m^3 + 54 n} + m)^2} {3 \sqrt[3]{6 \sqrt{81 n^2 - 3m^3n} - m^3 + 54 n}}\\
\shoveleft \text{When }m=1+2+5, n=1*2*5, s=10, 3\pm\sqrt{5} \text{ (removed)}\\
\shoveleft \implies [ABC]=10 \implies [DEF]=10-8=\bbox[5px, border: 1px solid black]{2}
\end{multline}\)
10/15/2024
$M$ is the midpoint of side $BC$ of $\triangle{ABC}$, and $D$ is a point on the circumcircle of $\triangle{ABC}$ such that $\angle{MAC}=\angle{BAD}$. $X,Y,Z$ are perpendicular feet from $D$ to $BC, AC, AB$ respectively. Prove: $XY=XZ$.
Prove:
\(\begin{multline}
\shoveleft AD \text{ is the A-symmedian of } \triangle{ABC} \implies \dfrac{AB}{AC}=\dfrac{BD}{CD}=\dfrac{sin\angle{ACB}}{sin\angle{ABC}}\\
\shoveleft \implies BD \cdot sin{\angle{ABC}}=CD \cdot sin{\angle{ACB}}\\
\shoveleft DZ\perp BZ, DX \perp BX \implies BXDZ \text{ is cyclic }, BD \text{ is the diameter of }(BXDZ)\\
\shoveleft \implies ZX=DZ \cdot sin\angle{XDZ} = BD \cdot sin\angle{ABC}\\
\shoveleft DX \perp XC, DY \perp YC \implies CDXY \text{ is cyclic }, CD \text{ is the diameter of }(CDXY)\\
\shoveleft \implies XY = CD \cdot sin{\angle{XCY}}= CD \cdot sin{\angle{ACB}}= BD \cdot sin\angle{ABC} = ZX \blacksquare\\
\end{multline}\)
Note: This proof is given here.
Prove 2:
\(\begin{multline}
\shoveleft DX \perp XC, DY \perp YC \implies CDXY \text{ is cyclic }\\
\shoveleft \implies \angle{MAC}=\angle{BAD}=\angle{XYD}, \angle{ACM}=\angle{XDY}\\
\shoveleft \implies \triangle{ACM}\sim\triangle{YDX}\implies \dfrac{XY}{AM}=\dfrac{XD}{CM}\\
\shoveleft DZ\perp BZ, DX \perp BX \implies BXDZ \text{ is cyclic }\\
\shoveleft \implies \angle{DAC}=\angle{MAC}+\angle{DAM}=\angle{BAD}+\angle{DAM}\\
\shoveleft =\angle{BAM}=\angle{DBC}=\angle{DZX}, \angle{XDZ}=\angle{ABM}\\
\shoveleft \implies \triangle{ABM}\sim\triangle{ZXD}\implies \dfrac{ZX}{AM}=\dfrac{XD}{BM}\\
\shoveleft =\dfrac{XD}{CM}=\dfrac{XY}{AM}\implies XD=XY\blacksquare
\end{multline}\)