Points $D$ and $E$ divides the base $BC$ of an isosceles triangle $ABC$ into three equal parts and $D$ is between $B$ and $E.$ Show that $\angle BAD<\angle DAE.$
Problem
Source: Finland 2011, Problem 3
Tags: trigonometry, geometry, symmetry, trig identities, Law of Sines, geometry unsolved
06.05.2013 00:39
Use the law of sines in $\triangle BAD$ and $\triangle DAE$. We have $\sin {\angle BAD}/BD=\sin {\angle DBA}/AD$ and $\sin \angle{DAE} /DE=\sin{\angle AED}/AD$. Dividing the two expressions we get $\sin {\angle BAD}/\sin{\angle DAE}=\sin{\angle DBA}/\sin{\angle AED}$. But $\angle AED>\angle ACB=\angle DBA$ and thus $\sin\angle{BAD}<\sin\angle{DAE}$. Because both angles are less than $90^\circ$ we get $\angle BAD <\angle DAE$. An alternative way would be to use the areas of $\triangle BAD$ and $\triangle DAE$, which are equal. But then $(AB)(AD)\sin\angle BAD=(AD)(AE)\sin\angle DAE$ and because $(AB)>(AE)$ the result follows.
06.05.2013 00:57
let $A'$ be opposite of $A$ wrt $D$ $\angle AEC>90>\angle ACE$ so $AE<AC=AB$ but $AB=EA'$ by symmetry so $EA'>AE$ now by symmetry $\angle BAD=\angle EA'D<\angle EAD$
27.03.2017 18:07
tk1 wrote: Use the law of sines in $\triangle BAD$ and $\triangle DAE$. We have $\sin {\angle BAD}/BD=\sin {\angle DBA}/AD$ and $\sin \angle{DAE} /DE=\sin{\angle AED}/AD$. Dividing the two expressions we get $\sin {\angle BAD}/\sin{\angle DAE}=\sin{\angle DBA}/\sin{\angle AED}$. But $\angle AED>\angle ACB=\angle DBA$ and thus $\sin\angle{BAD}<\sin\angle{DAE}$. Because both angles are less than $90^\circ$ we get $\angle BAD <\angle DAE$. An alternative way would be to use the areas of $\triangle BAD$ and $\triangle DAE$, which are equal. But then $(AB)(AD)\sin\angle BAD=(AD)(AE)\sin\angle DAE$ and because $(AB)>(AE)$ the result follows. how can you say angle DAE is acute??
29.03.2017 00:48
29.03.2017 00:52
My solution is quite similar to Trumpeter, except with a cosine monotonically increasing over that interval. Is this problem difficult enough for a #3?
29.03.2017 01:39
@Above: perhaps it is difficult enough for a Finland #3. What country are you basing your difficulty rating on?
29.03.2017 16:11
Let $ABA'E$ be a parallelogram; since clearly $A'B=AE<AB$, from $\triangle ABA'$ we get $\angle BAD<\angle BA'D=<DAE$, done. Best regards, sunken rock
07.08.2017 07:23
Sorry to revive an old(ish) thread, but how can we say that $(AB) > (AE)$. I have a decent diagram, and intuitively I agree with this statement, but could someone please provide a proof for this? Thank you!
07.08.2017 08:12
I'm not sure if this is a rigorous proof, but you can draw circle $A$ such that it passes through $B$ and $C$, and $E$ would be within the circle. This means $AE$ must be less than $AB$.