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

Latest revision as of 20:43, 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

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

Latest revision as of 20:43, 21 June 2021

Problem 26

Solution

See Also