AC = BD in convex ABCD, <A=<B=60^o,<CAB =<CBD (Kharkiv City VIII 2017- Ukr)

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AC = BD in convex ABCD, <A=<B=60^o,<CAB =<CBD (Kharkiv City VIII 2017- Ukr)

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#1 h Mar 19, 2020, 2:41 PM

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Let $ABCD$ be a convex quadrangle with $\angle A = \angle B = 60^o$ and $\angle CAB = \angle CBD$ . Prove that $AC = BD$ .

This post has been edited 1 time. Last edited by parmenides51, Mar 19, 2020, 9:11 PM

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brainiacmaniac31

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brainiacmaniac31

#2 h Mar 19, 2020, 6:12 PM

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Let $E=AD\cap BC$ , so that $\triangle ABE$ is equilateral.
$[asy] import olympiad; import geometry; pair a,b,c,d,f; a=dir(210); b=dir(-30); f=dir(90); c=(5f+2b)/7; d=(5a+2f)/7; dot("$A$",a,dir(a)); dot("$B$",b,dir(b)); dot("$C$",c,dir(c)); dot("$D$",d,dir(d)); dot("$E$",f,dir(f)); draw(a--b--f--a,linewidth(1.1)); draw(a--c^^b--d); draw(anglemark(b,a,c,5)); draw(anglemark(c,b,d,5)); [/asy]$
Note that by ASA congruence, $\triangle ABC\cong\triangle BED$ . Therefore, $AC=BD$ .

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