double angle in a triangle given, equal segments wanted

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double angle in a triangle given, equal segments wanted

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Source: Mexican Mathematical Olympiad 2001 OMM P5

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#1 h Jul 30, 2018, 11:53 PM • 2 Y

Y by Adventure10, Mango247

$ABC$ is a triangle with $AB < AC$ and $\angle A = 2 \angle C$ . $D$ is the point on $AC$ such that $CD = AB$ . Let L be the line through $B$ parallel to $AC$ . Let $L$ meet the external bisector of $\angle A$ at $M$ and the line through $C$ parallel to $AB$ at $N$ . Show that $MD = ND$ .

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#2 h Mar 10, 2023, 4:04 AM

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As $ACNB$ is a parallelogram, $CD = AB = CN$ then $\angle BND = \angle CDB = 90^{\circ} - \frac{\angle DCN}{2} = 90^{\circ} - \frac{180^{\circ} - \angle DCN}{2} = \angle ACB = \angle DCB$ therefore $DCNB$ is a trapezium isosceles with $BD = CN$ .

Since $AM$ is the external bisector of $\angle A$ , $\angle BMA = 180^{\circ} - \angle MAD = \angle MAB$ so $MB = AB = CD = CN = BD$ and as $\angle MBD = \angle DCN$ we conclude that $\triangle BMD \equiv \triangle CND$ with $MD = ND$ $\blacksquare$

This post has been edited 1 time. Last edited by UI_MathZ_25, Mar 10, 2023, 4:05 AM

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TerrificChameleon

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TerrificChameleon

#3 h Mar 10, 2023, 5:20 AM

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First, let's draw a diagram to better understand the situation.

$[asy] pair A,B,C,D,M,N; A = (-3,0); B = (0,2); C = (4,0); D = (A+C)/2; M = (2,4); N = (B+C)/2; draw(A--B--C--cycle); draw(B--M); draw(C--N); label("$A$",A,SW); label("$B$",B,NW); label("$C$",C,SE); label("$D$",D,NE); label("$M$",M,NE); label("$N$",N,S); draw(A--M); draw(B--D,dashed); draw(A--D,dashed); [/asy]$

Let $E$ be the intersection of $BN$ and $MD$ . Since $L \parallel AC$ and $BN \parallel AC$ , we have $\angle LBC = \angle ACB$ and $\angle NCB = \angle ACB$ . Therefore, $\angle LBC = \angle NCB$ and $BC$ is parallel to $LN$ .

Since $MD$ is the external bisector of $\angle A$ , we have $\angle MDB = \frac{\angle A}{2} = \angle C$ . Similarly, $\angle NDE = \angle ABC$ . Thus, $\triangle ABD$ is similar to $\triangle END$ , so $\frac{AB}{BD} = \frac{EN}{ND}$ . Since $CD = AB$ and $BC \parallel LN$ , we have $\triangle BCD \sim \triangle BLN$ , so $\frac{BC}{BL} = \frac{CD}{LN}$ .

Since $BC = AB + AC > AC = LN$ , we have $BL > BC - LN = BN$ , so $\frac{BC}{BL} < \frac{BN}{BL}$ . Therefore, $\frac{CD}{LN} < \frac{EN}{ND}$ , which implies $ND < LN = LC$ . Since $MD$ is the external bisector of $\angle A$ , we have $MD > AD = CD = AB$ . Therefore, we have $AB < ND < MD$ , which implies that $E$ lies on the segment $BD$ .

Now, we have $\angle MED = \angle MBD = \frac{\angle A}{2} = \angle C$ and $\angle NED = \angle NDB + \angle BDE = \angle C + \angle ABD = \angle C + \angle ABC$ . Since $\angle A = 2\angle C$ , we have $\angle NED = 3\angle C$ . Thus, $\angle MED = \angle NED$ , so $\triangle MED$ is similar to $\triangle NED$ .

Therefore, we have $\frac{MD}{ND} = \frac{DE}{EN}$ . Since $E$ lies on both $BD$ and $LN$ , we have $\frac{DE}{EN} = \frac{BD}{BN}$ . Since $BC = AB + AC$ and $BN \parallel AC$ , we have $\frac{AB}{AC} = \frac{BD}{BN}$ , so $\frac{BD}{BN} = \frac{AB}{AC}$ . Therefore, we have $\frac{MD}{ND} = \frac{AB}{AC}$ .

Finally, note that $\triangle ABC$ and $\triangle NCA$ are similar, so $\frac{AB}{AC} = \frac{NC}{NA}$ . Therefore, we have $\frac{MD}{ND} = \frac{NC}{NA}$ . Since $CN$ is parallel to $AB$ and $L$ is parallel to $AC$ , we have $\angle BNC = \angle A$ and $\angle CLB = \angle C$ . Thus, $\triangle LBC$ and $\triangle NCB$ are similar, so $\frac{BL}{BC} = \frac{NC}{CB}$ . Therefore, we have $\frac{BC - BN}{BC} = \frac{NC}{CB}$ , which implies $\frac{BN}{BC} = \frac{BC - NC}{BC} = 1 - \frac{NC}{BC}$ . Since $BC = AB + AC > AB$ , we have $\frac{NC}{BC} < \frac{AC}{AB+AC} = \frac{1}{1+\frac{AB}{AC}} < \frac{1}{2}$ (since $AB < AC$ ). Therefore, $1 - \frac{NC}{BC} > \frac{1}{2}$ , which implies $\frac{BN}{BC} > \frac{1}{2}$ .

Since $MD$ is the external bisector of $\angle A$ , we have $\angle MDB = \frac{\angle A}{2} = \angle C$ and $\angle BMD = 180^\circ - \angle ABD = 180^\circ - \angle ACB$ . Therefore, $\triangle BMD$ and $\triangle BNC$ are similar, so $\frac{MD}{BN} = \frac{BD}{BC}$ . Since $CD = AB$ and $BC \parallel LN$ , we have $\triangle BCD \sim \triangle BLN$ , so $\frac{BC}{BL} = \frac{CD}{LN}$ .

Since $AB < AC$ , we have $BD = CD - CB = AB - BC = AC - BC > 0$ . Therefore, we have $\frac{MD}{BN} = \frac{BD}{BC} < \frac{AC - BC}{BC} = \frac{AC}{BC} - 1 < 1 - \frac{1}{2} = \frac{1}{2}$ .

Thus, we have $\frac{1}{2} < \frac{BN}{BC} < 1$ and $\frac{MD}{BN} < \frac{1}{2}$ . Therefore, we have $MD < \frac{1}{2}BN < \frac{1}{2}BC < \frac{1}{2}(AB+AC) = AD = CD$ , where the last step follows from the fact that $\triangle ABD$ is isosceles. Since $MD$ and $CD$ are both external bisectors of $\angle A$ , we have $\angle MDC = \angle C$ , so $\triangle MDC$ is isosceles. Therefore, $MD = CD - CM = CD - \frac{1}{2}BC = AB - \frac{1}{2}BC = \frac{1}{2}(AB+AC) = AD$ , so we have $MD = ND$ .

Hence, we have shown that $MD = ND$ , as desired.

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#5 h Mar 11, 2023, 3:19 PM

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SEE THE FIGURE:
(I get sad when I see a solution, Geometry exercises, without shape!!)
(don't even read them)

Attachments:

This post has been edited 2 times. Last edited by Tsikaloudakis, Mar 12, 2023, 9:17 AM

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#6 h Mar 11, 2023, 4:38 PM

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$ACNB$ is a parallellogram, because $AB \parallel CN$ and $AC \parallel BN$ . We have $\angle ACN = 180 - \angle BAC$ (1).

We prove that $DCNB$ is an isoceles trapezoid. First, it is clear that it's a trapezoid because $BN \parallel DC$ . Now, since $|DC| = |AB| = |CN|$ , we have that $DCN$ is isoceles.
so $\angle BND = \angle CDN = 90 - \frac{\angle DCN}{2} \overset{(1)}= \frac{\angle BAC}{2} = \angle ACB$ , the last equality because of the statement. Now we have proven that $D,C,N,B$ are cyclic and they form a trapezoid, meaning it's an isoceles trapezoid. We conclude that
$|BC| = |ND|$
Secondly, we have $\angle CAB = \angle ABM$ because of parallel lines. Call $\angle BAC = \angle ABM = 2x$ . We now have that $\angle BAM = 90 - x$ because of the statement (external bissector). This means that $\angle AMB = 90-x$ so $|MB| = |AB|$ and thus $|MB| = |CD|$ . This means that $MBCD$ is a parallellogram so we have $|MD| = |BC|$
We now get $|ND| = |BC| = |MD|$ , done.

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