Let ABC be a triangle with ∠A=90∘ and AB=AC. Let D and E be points on the segment BC such that BD:DE:EC=1:2:√3. Prove that ∠DAE=45∘
Problem
Source: Problem 4
Tags: geometry, Angle Chasing, similar triangles
11.12.2013 07:35
There are a lot of ways to solve it and here is one of them. Let BD=2,DE=4,EC=2√3 and we obtain AB2=6(√3+1)2=(BE+DE)⋅(DE+EC)=BE⋅DC,and hence ABBE=DCCA,together with ∠ABE=∠DCA=45∘.It follows that △ABE∼△DCA and therefore,∠EAB=∠BAD+∠DAE=∠CDA=∠BAD+∠ABC⟹∠DAE=45∘,as desired. Q.E.D.
23.02.2014 20:36
For a generalization see RMO 2003 problem 1. Bye
23.09.2015 18:27
Just apply cosine law in DAC & in BAE and then after getting all three sides of EAD, apply cosine law in EAD to get DAE=45°.
22.11.2015 09:40
Rotating the configuraiton about Aby 90◦, the point B goes to the point C. Let P denote the image of the point D under this rotation. Then CP=BD and ∠ACP =∠ABC = 45◦, so ECP is a right-angled triangle with CE:CP = √3:1. Hence PE=ED. It follows that ADEP is a kite with AP=AD and PE=ED. Therefore AE is the angular bisector of ∠PAD. This implies that ∠DAE = ∠PAD/2 = 45◦
22.11.2015 09:49
aayush-srivastava wrote: Rotating the configuraiton about Aby 90◦, the point B goes to the point C. Let P denote the image of the point D under this rotation. Then CP=BD and ∠ACP =∠ABC = 45◦, so ECP is a right-angled triangle with CE:CP = √3:1. Hence PE=ED. It follows that ADEP is a kite with AP=AD and PE=ED. Therefore AE is the angular bisector of ∠PAD. This implies that ∠DAE = ∠PAD/2 = 45◦ Can you explain a bit more plz..
22.08.2016 07:07
If we use sine law in this problem , will we get full marks ?
01.10.2018 05:22
Yes of course we will get full marks