In the figure below, we know that AB=CD and BC=2AD. Prove that ∠BAD=30o.
Problem
Source: IGO 2015 Elementary 3
Tags: geometry, Elementary, angles
22.07.2018 16:11
Let the foot of the altitudes from D to BC be Q, and let the midpoint of AB be M. We have BM=12AB=12CD=PD. Since BM||PD, this implies that BMPD is a rectangle, and MD⊥BM. We now see that MD is the perpendicular bisector AB, so AD=BD. Finally, we have BC=2AD=2BD, and by Sine Law, we can get that ∠BDC=90∘, and therefore, ∠DBC=60∘. Hence, ∠BAD=∠ABD=90∘−∠DBC=30∘.
22.07.2018 17:44
Construct the rectangle ABCE, see that △CDE is equilateral, while in △DAE,AE=2AD and m(^AED)=30∘, that is, ^DAE=60∘, or ^BAD=30∘. Best regards, sunken rock
20.09.2022 12:42
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20.09.2022 15:56
This is the first time I was and probably last time I will be, able to solve a Geo doing a construction. Construct a rectangle ABCD′ such that AB=CD′ and BC=AD′ then by some simple angle chase we get ∠DD′A=30o. Let ∠BAD=x⟹∠DAD′=90−x and ∠ADD′=60+x. Since BC=AD′=2AD , Using sine law we get sin(60+x)2AD=sin(30o)AD⟹x=30osince clearly ∠BAD is acute.
04.12.2023 22:57
Extend CD to meet AB at P So, angle CPB=60° Let AD=x, so, BC=2x In ∆ BCP, CP =2BP PD+CD=2BP PD+AB=2BP( CD=AB) PD+AP+PB=2BP AP+PD=BP In∆BCP, tan 60°=2x/BP BP/x=2/√3 Let angle BAD=y angle PDA=60°-y So, by laws of sine, PD+AP/ AD={sin y+ sin(60°-y)}/ sin 60° BP/x={sin y+ sin(60°-y)}/ sin 60° {sin y+ sin(60°-y)}/ sin 60°=2/√3 sin y +√3 cos y=2 4cos²y-4√3cos y+3=0 (2cos y-√3)²=0 cos y=√3/2 y=30° angle BAD=30° @Krishijivi