MathDB

3

Part of 1986 AMC 12/AHSME

Problems(1)

Easy chapter level geometry

Source:

10/26/2005
△ABC\triangle ABC is a right angle at CC and ∠A=20∘\angle A = 20^\circ. If BDBD is the bisector of ∠ABC\angle ABC, then ∠BDC=\angle BDC = [asy] size(200); defaultpen(linewidth(0.8)+fontsize(11pt)); pair A= origin, B = 3 * dir(25), C = (B.x,0); pair X = bisectorpoint(A,B,C), D = extension(B,X,A,C); draw(B--A--C--B--D^^rightanglemark(A,C,B,4)); path g = anglemark(A,B,D,14); path h = anglemark(D,B,C,14); draw(g); draw(h); add(pathticks(g,1,0.11,6,6)); add(pathticks(h,1,0.11,6,6)); label("AA",A,W); label("BB",B,NE); label("CC",C,E); label("DD",D,S); label("20∘20^\circ",A,8*dir(12.5)); [/asy]
<spanclass=′latex−bold′>(A)</span> 40∘<spanclass=′latex−bold′>(B)</span> 45∘<spanclass=′latex−bold′>(C)</span> 50∘<spanclass=′latex−bold′>(D)</span> 55∘<spanclass=′latex−bold′>(E)</span> 60∘ <span class='latex-bold'>(A)</span>\ 40^\circ \qquad <span class='latex-bold'>(B)</span>\ 45^\circ \qquad <span class='latex-bold'>(C)</span>\ 50^\circ \qquad <span class='latex-bold'>(D)</span>\ 55^\circ \qquad <span class='latex-bold'>(E)</span>\ 60^\circ
geometryLaTeX