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Difference between revisions of "1975 AHSME Problems/Problem 26"

(Created page with "== Problem 26 == In acute <math>\triangle ABC</math> the bisector of <math>\measuredangle A</math> meets side <math>BC</math> at <math>D</math>. The circle with center <math>...")
 
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\textbf{(D)}\ AM-AN=\frac{3(DB-DC)}{2}\qquad
 
\textbf{(D)}\ AM-AN=\frac{3(DB-DC)}{2}\qquad
 
\textbf{(E)}\ AB-AC=\frac{3(DB-DC)}{2}  </math>
 
\textbf{(E)}\ AB-AC=\frac{3(DB-DC)}{2}  </math>
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==Solution==
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Error
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==See Also==

Latest revision as of 19:43, 21 June 2021

Problem 26

In acute $\triangle ABC$ the bisector of $\measuredangle A$ meets side $BC$ at $D$. The circle with center $B$ and radius $BD$ intersects side $AB$ at $M$; and the circle with center $C$ and radius $CD$ intersects side $AC$ at $N$. Then it is always true that

$\textbf{(A)}\ \measuredangle CND+\measuredangle BMD-\measuredangle DAC =120^{\circ} \qquad \textbf{(B)}\ AMDN\ \text{is a trapezoid}\qquad \textbf{(C)}\ BC\ \text{is parallel to}\ MN\\ \qquad \textbf{(D)}\ AM-AN=\frac{3(DB-DC)}{2}\qquad \textbf{(E)}\ AB-AC=\frac{3(DB-DC)}{2}$

Solution

Error

See Also

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