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In the figure below, $D$ is a point on segment $\overline{CE}$ such that $\overline{AD}\parallel\overline{BE}$ and $A$ is not on $\overline{BC}.$


Line segments $\overline{AD}$ and $\overline{BC}$ intersect at $P.$  [asy] size(4cm); pair C = (0,0); pair D = (4,0); pair E = (9,0); pair F = (0.25,4); pair G = (1,4); pair A = extension(C,F,D,G); pair H = G+E-D; pair B = extension(C,G,E,H); pair P = extension(A,D,B,C);  draw(A--C--E--B--C); draw(A--D);  label("$A$",A,NNW); label("$B$",B,N); label("$C$",C,SSW); label("$D$",D,S); label("$E$",E,SSE); label("$P$",P,dir(0)); [/asy]  

 

 

Can $\angle CAD=\angle CBE?$ Explain.

 Aug 26, 2020
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Yes, angle CAD can be equal to CBE, by placing points D and B close enough together.

 Aug 27, 2020

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