Random quick geo lemmas

by Inconsistent, Sep 26, 2021, 1:22 AM

Ever see the condition come up $D \in AB, E \in AC, BD = x, CE = y$? Not sure what it means? Well one way to interpret it is that the center of spiral similarity from the segment $BC$ to $DE$ is the point $X$ on the circumcircle of $\triangle ABC$ such that $\frac{BX}{XC} = \frac{x}{y}$. This comes from $\triangle BDX \sim \triangle CEX$ after $\angle DBX = \angle ABX = ACX = \angle ECX$.
[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(0cm); unitsize(30);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -10.457861904816939, xmax = 4.966382807294912, ymin = -4.336543019812374, ymax = 6.20830120164101;  /* image dimensions */
pen qqwuqq = rgb(0,0.39215686274509803,0); 

draw(arc((-6.34,-0.21),0.4449301359263034,50.82615299044904,66.42867268465027)--(-6.34,-0.21)--cycle, linewidth(1) + qqwuqq); 
draw(arc((3.22,-0.23),0.4449301359263034,128.93411606425818,144.53663575845943)--(3.22,-0.23)--cycle, linewidth(1) + qqwuqq); 
 /* draw figures */
draw(circle((-1.5578941209700194,0.7866101763305489), 4.884850917051483), linewidth(1)); 
draw((-6.34,-0.21)--(3.22,-0.23), linewidth(1)); 
draw(circle((-2.0119933619256116,2.425966215820826), 3.2785300593657425), linewidth(1)); 
draw((-5.283503068252974,2.2115283656856604)--(-6.34,-0.21), linewidth(1)); 
draw((-5.730190237421473,0.965179524434851)--(-5.8933128308315,1.0363488412508102), linewidth(1)); 
draw((-5.283503068252974,2.2115283656856604)--(-4.08,4.97), linewidth(1)); 
draw((-4.08,4.97)--(1.068153770457503,1.302821971728902), linewidth(1)); 
draw((1.068153770457503,1.302821971728902)--(3.22,-0.23), linewidth(1)); 
draw((2.1957050013407637,0.608888918098619)--(2.092448769116741,0.46393305363028287), linewidth(1)); 
draw((-6.34,-0.21)--(-1.5476747899503867,5.671450403716929), linewidth(1)); 
draw((-4.036243246851684,2.7581917891858714)--(-3.8982734547739297,2.645771201165327), linewidth(1)); 
draw((-3.989401335176457,2.815679202551603)--(-3.851431543098703,2.7032586145310584), linewidth(1)); 
draw((-1.5476747899503867,5.671450403716929)--(3.22,-0.23), linewidth(1)); 
draw((0.8820816009049207,2.8054878156560346)--(0.7436426395488169,2.6936454886259376), linewidth(1)); 
draw((0.9286825705007941,2.747804915090992)--(0.7902436091446904,2.6359625880608952), linewidth(1)); 
draw((-5.283503068252974,2.2115283656856604)--(-1.5476747899503867,5.671450403716929), linewidth(1)); 
draw((1.068153770457503,1.302821971728902)--(-1.5476747899503867,5.671450403716929), linewidth(1)); 
 /* dots and labels */
dot((-4.08,4.97),dotstyle); 
label("$A$", (-4.021205938416417,5.125637870887006), NW * labelscalefactor); 
dot((-6.34,-0.21),dotstyle); 
label("$B$", (-6.275518627109687,-0.06521371491986393), SW * 2); 
dot((3.22,-0.23),dotstyle); 
label("$C$", (3.275648290774959,-0.0800447194507407), SE * labelscalefactor); 
dot((-5.283503068252974,2.2115283656856604),dotstyle); 
label("$D$", (-5.222517305417436,2.3670710281439264), SW *2); 
dot((1.068153770457503,1.302821971728902),linewidth(4pt) + dotstyle); 
label("$E$", (1.1251526337978262,1.4178867381678133), E * labelscalefactor); 
dot((-1.5476747899503867,5.671450403716929),linewidth(4pt) + dotstyle); 
label("$H$", (-1.4851041636364872,5.793033074776461), N * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]

Here's another random one: If cevians $AD, BE, CF$ concur, then sometimes we will wonder about properties of $EF \cap (ABC) = P, Q$ on the sides of $B, C$ respectively. We know $PQ$ swap under the nice inversion at the harmonic conjugate of $D$ on $BC$ when the concurrence is due to Miquel (meaning $(BCEF), (ABDE), (ACDF)$). Here's another one: if $Q' = DQ \cap (ABC), Q'' = AQ' \cap EF$, then $(P, Q''; E, F)$ is a harmonic bundle. This is by projecting the ceva-menelaus on $BC$ via $E$ onto $(ABC)$, then projecting via $A$ onto $EF$.

Another thing that isn't really a lemma, when there is an arbitrary point on $BC$ that is the sole degree of freedom beyond the triangle, it might be a good idea, in general, to consider its projection via $A$ onto $(ABC)$ when experimenting for similar triangles and cyclic quadrilaterals.

Comment

3 Comments

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gj improving in geo ;)

thx, now I can shift all the blame to nt instead hehe ;)
This post has been edited 1 time. Last edited by Inconsistent, Sep 26, 2021, 6:08 AM
Reason: edit

by kevinmathz, Sep 26, 2021, 4:13 AM

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omg hi :love:

by nukelauncher, Sep 28, 2021, 4:20 AM

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hi :blush:

by AwesomeYRY, Nov 8, 2021, 2:47 AM

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