On $\triangle ABC$ let points $D,E$ on sides $AB,BC$ respectivily such that $AD=DE=EC$ and $AE \ne DC$. Let $P$ the intersection of lines $AE, DC$, show that $\angle ABC=60$ if $AP=CP$.
Problem
Source: Bolivia Cono Sur TST 2022 Day 2 P6
Tags: geometry, Bolivia, TST, Spiral Similarity, Angle Chasing
kante314
28.07.2022 09:27
Let $\angle DAE=\angle DEA=a$ and $\angle CAE=\angle PCA=b.$ Then, we know $\angle APC=180-2b,$ so $\angle CPE=2b.$ We also know $\angle APD=2b.$ Therefore, $\angle ADP=180-2b-a$ and $\angle PEC=180+a-4b.$
Now, by Sine Law on $\triangle ADP$ and $\triangle CPE,$ we have
$$\frac{\sin 180-2b-a}{AP}=\frac{\sin 2b}{AD} \text{ and } \frac{\sin 2b}{CE}=\frac{\sin 180+a-4b}{CP}$$Since $AD=CE$ and $AP=CP,$ we see $\sin 180-2b-a=\sin 180+a-4b.$ So, we have the two cases:
Case 1: $180-2b-a=180+a-4b$
This means $\angle ADE \cong \angle CED,$ so $\triangle ADE \cong \triangle CED.$ However, $AE\neq DC,$ so this isn't possible.
Case 2: $2b+a=180+a-4b$
This means $6b=180$ and $b=30.$ Thus, $$\angle ABC = 180-(a+b)-(3b-a)=180-4b=60,$$as desired.
samrocksnature
28.07.2022 09:36
kante314 wrote:
SolutionLet $\angle DAE=\angle DEA=a$ and $\angle CAE=\angle PCA=b.$ Then, we know $\angle APC=180-2b,$ so $\angle CPE=2b.$ We also know $\angle APD=2b.$ Therefore, $\angle ADP=180-2b-a$ and $\angle PEC=180+a-4b.$
Now, by Sine Law on $\triangle ADP$ and $\triangle CPE,$ we have
$$\frac{\sin 180-2b-a}{AP}=\frac{\sin 2b}{AD} \text{ and } \frac{\sin 2b}{CE}=\frac{\sin 180+a-4b}{CP}$$Since $AD=CE$ and $AP=CP,$ we see $\sin 180-2b-a=\sin 180+a-4b.$ So, we have the two cases:
Case 1: $180-2b-a=180+a-4b$
This means $\angle ADE \cong \angle CED,$ so $\triangle ADE \cong \triangle CED.$ However, $AE\neq DC,$ so this isn't possible.
Case 2: $2b+a=180+a-4b$
This means $6b=180$ and $b=30.$ Thus, $$\angle ABC = 180-(a+b)-(3b-a)=180-4b=60,$$as desired.
OlympusHero
29.07.2022 07:14
I did it quite similarly to kante314; the only difference is that I had $b = \angle CDE$. This made the equation $2a+b=180-a-2b \implies a+b= 60$ and via angle chasing $\angle ABC = 180-2(a+b) = 60$ as desired.