if <B = 2< C$, AD is altitude, BM=MC, prove AB = 2DM

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if <B = 2< C$, AD is altitude, BM=MC, prove AB = 2DM

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Source: V.A. Yasinsky Geometry Olympiad 2018 VIII-IX p3, advanced p2 [Ukraine]

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#1 h Oct 1, 2018, 6:48 PM • 2 Y

Y by Adventure10, Mango247

In the triangle $ABC$ , $\angle B = 2 \angle C$ , $AD$ is altitude, $M$ is the midpoint of the side $BC$ . Prove that $AB = 2DM$ .

This post has been edited 3 times. Last edited by parmenides51, Nov 7, 2022, 9:30 PM

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#2 h Oct 2, 2018, 3:21 PM • 1 Y

Y by Adventure10

Let $E$ be reflection of $A$ in perpendicular bisector of $BC$ ; easily $ABCE$ is an isosceles trapezoid with $CE=AE=AB$ ; if $F$ is projection of $E$ onto $BC$ , $ADFE$ is a rectangle, hence $AB=AE=DF=2DM$ , done.

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#3 h Oct 2, 2018, 4:04 PM • 1 Y

Y by Adventure10

Construct angle bisector $BN$ such that $\angle NBC=\angle NCB=\angle B$ . $NBC$ is a isosceles triangle so $NM$ is perpendicular bisector of $\angle BNC$ and $NM \parallel AD$ . By thales theorem $AN/NC=DM/MC$ And by angle bisector theorem $AN/NC=AB/BC$ So ultimately $AB/BC=DM/MC$ or $AB=2DM$ . so done

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#4 h Nov 3, 2018, 8:23 PM • 2 Y

Y by Adventure10, Mango247

Let $AB=x \text{ and } \angle B = 2y= 2\angle C$ ; Hence, $BD=x\cos 2y \text{ and } DC=2x\cos ^2y \implies BM= \frac{x}{2}[\cos 2y +2\cos ^2y] \implies DM=\frac{x}{2}[\cos 2y +2\cos ^2y] - x\cos 2y = \frac{x(2\cos ^2y -2\cos ^2y +1)}{2}=\frac{x}{2}=\frac{1}{2}AB \implies \boxed{AB=2DM}$

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#5 h Nov 4, 2018, 4:48 AM • 2 Y

Y by Adventure10, Mango247

Let $N$ bet middle of $AB$ . Let $\angle NBD=2x, \angle BCD=x$ . Then $\angle BDN=2x$ ,this why $\angle NDA=90-2x$ .We must prove that $ND=DM$ , or $\angle DMN=x$ . We know that $NM||AC$ , this why $,\angle DMN=\angle DNM =x$ , and $2DM=AB$

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#6 h Aug 26, 2023, 9:23 PM • 1 Y

Y by FriIzi

Midline !!

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