angle wanted, 105-45-30 triangle, median (2016 Moldova NMO 8.3)

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angle wanted, 105-45-30 triangle, median (2016 Moldova NMO 8.3)

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#1 h Mar 19, 2021, 3:34 PM

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Let the triangle $ABC$ be with $\angle B=105^o$ , $\angle C = 30^o$ . If $D \in (BC)$ so that $[AD]$ is the median of the triangle, to calculate the measure of the angle $DAC$ .

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#2 h Mar 19, 2021, 5:47 PM • 3 Y

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Solution

This post has been edited 6 times. Last edited by Euclides, Mar 20, 2021, 12:51 AM

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#3 h Mar 19, 2021, 6:52 PM

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This post has been edited 2 times. Last edited by natmath, Mar 19, 2021, 6:53 PM

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#4 h Mar 19, 2021, 7:28 PM • 1 Y

Y by Mango247

parmenides51 wrote:

Let the triangle $ABC$ be with $\angle B=105^o$ , $\angle C = 30^o$ . If $D \in (BC)$ so that $[AD]$ is the median of the triangle, to calculate the measure of the angle $DAC$ .

Reflect $B$ about $AC$ to $B'$ . Then we have that the quadrilateral $ABCB'$ is a kite. So its diagonals are perpendicular i.e. $BB' \perp AC$ . So $\angle CBB' =90-\angle BCA=60$ . Hence we can also say that $\angle BB'C=60$ i.e. $\triangle BB'C$ is equilateral i.e $BB'=BC$ . Also note that $\angle BAB' =\angle BAC+\angle B'AC=2\angle BAC=90$ hence by Pythagoras theorem on $\triangle BAB'$ , $BC^2=BB'^2=BA^2+B'A^2=2BA^2$ so we have $BA^2=\frac {BC^2}{2}=BC\cdot BD$ . Hence we have that $BA$ is tangent to $(ADC)$ . So $\angle BAD=\angle ACD=30$ hence $\angle DAC=45-\angle BAD=\boxed{15}$

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