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a1eec082-0233-4140-9334-aa3e39e27e1b 142 Alvaro Diaz Falconi 4 5 false false false 0 0 decimal

Elementary triangle geometry

z# RE z# z# Conjugate + 2 / 11 line :

z# IM z# z# Conjugate - 2 i * / 11 line :

a b i * + RE a

a b i * + Re a b i * + Re

M# z2xy A# 0 : r# 0 : 12 mat a# M# sys2mat.1 r# r# 1 + : A# r# 1 el a# RE : A# r# 2 el a# IM : 13 mat for A# 31 line :

A# xy2z A# 1 el i A# 2 el * + 11 line :

Line A# B# gL R# A# IM B# IM - (z RE * A# RE B# RE - (z IM * - A# RE B# IM * A# IM B# RE * - (+ numden : R# 1 el R# 2 el * 21 line :

Intersection a# b# g∩ a# b# 21 mat x y 21 mat roots xy2z 11 line :

At x# A# g\0040\ x# x A# RE ≡ y A# IM ≡ 21 sys at 11 line :

Paralel at A to a a# A# g\007C\ a# a# A# g\0040\ - 11 line :

Parpendicular at A to a a# A# g⟂ a# i z * g\0040\ A# g\007C\ 11 line :

∥

Signed distance r# A# gδ r# A# g\0040\ r# x diff (2^r# y diff (2^+ sqrt / 11 line :

Signed bisector r# s# gβ s# x diff 2^s# y diff 2^+ sqrt r# * r# x diff 2^r# y diff 2^+ sqrt s# * + 11 line :

Circle center and radious from 3 points A# B# C# gC eq# x a# - (2^y b# - (2^+ r# 2^- : a# b# r# 31 mat 00031 mat u# f# a# b# r# 13 mat u# 1 el u# 2 el u# 3 el 13 mat : eq# A# g\0040\ eq# B# g\0040\ eq# C# g\0040\ 31 mat 21 line : al_nleqsolve : a# i b# * + r# 12 mat 31 line :

XYPlot shorthands S# A# B# gXY S# Circle ≡ circle A# Re A# Im B# 13 mat transpose 11 mat blue solid stack 11 mat 11 mat S# Point ≡ A# str2num z2xy B# 8 blue augment eval A# str2num z2xy A# 10 blue augment eval 21 sys S# Poly ≡ polygon A# str2num vectorize z2xy 11 mat # B# 4 substr concat solid 2 B# stack 11 mat 11 mat A# str2num vectorize z2xy A# transpose 10 black augment eval 21 sys Nothing else if 11 line :

Problem: construct a triangle from an angle A, its bisector line h and the opposite side. angle A (BAC) 40 degrees, opposite side a (BC) 4 cm and angle A's bisector h (AD) 3 cm

Ref: https://en.smath.com/forum/yaf_postst25461findunread_roots-function.aspx

z x i y * + :

clr #2F0FF004 :

α 40 deg * : B 00 i * + : C 40 i * + : ho 3 : 41 sys

Angle A
Point B
Point C
bisector length

Π M 0.5 B C + (* : Point M . gXY 21 line μ.B B C gL M g⟂ : tg.B α - tan x * y - : ptg.B tg.B B g⟂ : F μ.B ptg.B g∩ : Point F . gXY 21 line Arc.F F z - abs F B - abs - : u eq x u 1 el : y u 2 el : D u 3 el : 13 mat x i y * + D - abs ho - Arc.F x i y * + (B - abs x i y * + (C - abs / B D - abs C D - abs / - 31 mat 21 line : σ 111 stack eq al_nleqsolve : A σ 1 el i σ 2 el * + : Poly A B C 13 mat clr gXY 41 line D σ 3 el i 0 * + : Point D . gXY 21 line h A C gL A B gL gβ : 91 sys :

A 1.085 - 1.9784 i * +

A D - abs 3

h y diff h x diff atan deg 48.74

BT: Bisector theorem:
(from Wikipedia)

M 0.5 B C + (* : Point M . gXY 21 line μ.B B C gL M g⟂ : tg.B α - tan x * y - : ptg.B tg.B B g⟂ : F μ.B ptg.B g∩ : Point F . gXY 21 line Arc.F F z - abs F B - abs - : u eq x u 1 el : y u 2 el : D u 3 el : 13 mat x i y * + D - abs ho - Arc.F x i y * + (B - abs x i y * + (C - abs / B D - abs C D - abs / - 31 mat 21 line : σ 111 stack eq al_nleqsolve : A σ 1 el i σ 2 el * + : Poly A B C 13 mat clr gXY 41 line D σ 3 el i 0 * + : Point D . gXY 21 line h A C gL A B gL gβ : 91 sys

M: the midpoint of BC

μ.B: the perp at M of BC

tg.B: the line by B with slope α

ptg.B: the perp at B of tgB

F: μ.B ∩ ptg.B

Arc.F: the circle with center F and radious OB. It is the locus of the point A if the angle BAC = α (and this is why I call tg to the line with slope α)

eq: conditions over the point A(x,y) and D

σ: eq solution, returning xA,yA and xD

gXY draw the triangle

draw the point D

draw the bisector h

Alvaro

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