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We can view these conditions as a geometry diagram as seen below. So, we know that ef=34\frac{e}{f}=\frac{3}{4} (since e=a−b=34c−34d=34fe=a-b=\frac{3}{4} c-\frac{3}{4} d=\frac{3}{4} f and we know that e2+f2=15\sqrt{e^{2}+f^{2}}=15 (since this is a2+c2−b2+d2)\left.\sqrt{a^{2}+c^{2}}-\sqrt{b^{2}+d^{2}}\right). Also, note that ac+bd−ad−bc=(a−b)(c−d)=efa c+b d-a d-b c=(a-b)(c-d)=e f. So, solving for ee and ff, we find that e2+f2=225e^{2}+f^{2}=225, so 16e2+16f2=360016 e^{2}+16 f^{2}=3600, so (4e)2+(4f)2=3600(4 e)^{2}+(4 f)^{2}=3600, so (3f)2+(4f)2=3600(3 f)^{2}+(4 f)^{2}=3600, so f2(32+42)=3600f^{2}\left(3^{2}+4^{2}\right)=3600, so 25f2=360025 f^{2}=3600, so f2=144f^{2}=144 and f=12f=12. Thus, e=3412=9e=\frac{3}{4} 12=9. Therefore, \boldsymbol{e f}=\mathbf{9} * \mathbf{1 2}=\mathbf{1 0 8}$.
Plain-text mathematical notation (without MathML)
We can view these conditions as a geometry diagram as seen below. So, we know that (e)/(f)=(3)/(4) (since e=a−b=(3)/(4)c−(3)/(4)d=(3)/(4)f and we know that √(e²+f²)=15 (since this is √(a²+c²)−√(b²+d²)). Also, note that ac+bd−ad−bc=(a−b)(c−d)=ef. So, solving for e and f, we find that e²+f²=225, so 16e²+16f²=3600, so (4e)²+(4f)²=3600, so (3f)²+(4f)²=3600, so f²(3²+4²)=3600, so 25f²=3600, so f²=144 and f=12. Thus, e=(3)/(4)12=9. Therefore, \boldsymbol{e f}=\mathbf{9} * \mathbf{1 2}=\mathbf{1 0 8}$.
Original LaTeX notation
We can view these conditions as a geometry diagram as seen below. So, we know that $\frac{e}{f}=\frac{3}{4}$ (since $e=a-b=\frac{3}{4} c-\frac{3}{4} d=\frac{3}{4} f$ and we know that $\sqrt{e^{2}+f^{2}}=15$ (since this is $\left.\sqrt{a^{2}+c^{2}}-\sqrt{b^{2}+d^{2}}\right)$. Also, note that $a c+b d-a d-b c=(a-b)(c-d)=e f$. So, solving for $e$ and $f$, we find that $e^{2}+f^{2}=225$, so $16 e^{2}+16 f^{2}=3600$, so $(4 e)^{2}+(4 f)^{2}=3600$, so $(3 f)^{2}+(4 f)^{2}=3600$, so $f^{2}\left(3^{2}+4^{2}\right)=3600$, so $25 f^{2}=3600$, so $f^{2}=144$ and $f=12$. Thus, $e=\frac{3}{4} 12=9$. Therefore, \boldsymbol{e f}=\mathbf{9} * \mathbf{1 2}=\mathbf{1 0 8}$.

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