$ADC$ and $ABC$ are triangles such that $AD=DC$ and $AC=AB$. If $\angle CAB=20^{\circ}$ and $\angle ADC =100^{\circ}$, without using Trigonometry, prove that $AB=BC+CD$.
Problem
Source: NMTC 2017 Juniors P3
Tags: geometry
phymaths
28.10.2017 15:03
Easy ! Let $AD$ and $BC$ intersect at $P$ and let $I$ be the intersection of $CD$ and $AB$. Simple Angle chasing gives $I$ as in centre of $PAC$ and $PDIB$ is cyclic. Then take a point $X$ on $AC$ such that $CX=BC$. Then again by some angles we get $CBDX$ as a kite and also $AX=AD$ Thus $AB=BC+CD$
AnArtist
28.10.2017 15:04
assume D lies on opposite side of B. Extend BC till E so that CE = CD you will find that DCE is equilateral and then angle chasing gives BE=AB which is the desired. when D lies on the same side then the new triangle is congruent to previous and the result is still true.
soumyajit_math
02.05.2020 07:41
AnArtist wrote: assume D lies on opposite side of B. Extend BC till E so that CE = CD you will find that DCE is equilateral and then angle chasing gives BE=AB which is the desired. when D lies on the same side then the new triangle is congruent to previous and the result is still true. After facts are come from the isoscsles triangle ADE
OlympusHero
25.09.2021 05:43
using trig just to tick off the op
Denote $AC=x$. We have $AB=x, BC=\frac{x \sin 20}{\sin 80}, CD=\frac{x \sin 40}{\sin 100}$, so it suffices to show $\frac{\sin 20}{\sin 80}+\frac{\sin 40}{\sin 100} = 1$, or $\frac{\cos 80 + \cos 40 + 1}{\cos 20 + 1}$ after writing over a common denominator, using Product-to-Sum, and simplifying. Next, we have $\cos 80 + \cos 40 = \cos 20$ from Sum-to-Product, so $\frac{\cos 80 + \cos 40 +1}{\cos 20+1}=1$ as desired.