My Math Forum Problem with isosceles triangles

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 April 10th, 2015, 07:18 PM #1 Newbie   Joined: Sep 2014 From: Brazil Posts: 1 Thanks: 0 Problem with isosceles triangles Hi; this is my first post. I'm puzzling my brain with this dumb problem, without success . Here's the question: one big isosceles triangle (AB=AC) and another isosceles, inside the big one (AD=AE) – see the representation below. The angle ∠BAD is 20º. What the value of angle ∠EDC? https://imageshack.us/i/exbKvtpLj The alternatives for the problem are a) 5º b) 10º c) 15º d) 20º e) 25º I tried using the exterior angle theorem, but I got 20º; then I used ruler, compass and protractor and... 10º. What the... Hoping help... Last edited by Avlis Odraude; April 10th, 2015 at 07:35 PM.
April 10th, 2015, 08:33 PM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

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Hello, Avlis Odraude!

Welcome aboard!

Quote:
 One big isosceles triangle (AB=AC) and another isosceles, inside the big one (AD=AE).$\;$ See the representation below. The angle ∠BAD is 20º. What the value of angle ∠EDC? https://imageshack.us/i/exbKvtpLj The alternatives for the problem are $\;\;(a)\;5^o\;\;\;(b)\;10^o\;\;\;(c)\;15^o \;\;\;(d)\;20^o \;\;\;(e)\;25^o$

$\text{Let }\theta \,=\,\angle DAE.$

$\text{In isosceles }\Delta ABC:\:\angle B \,=\, \angle C \,=\,80\,-\,\frac{1}{2}\theta$

$\text{In isosceles }\Delta ADE:\: \angle ADE \,=\,\angle AED \,=\, 90\,-\,\frac{1}{2}\theta$
$\;\;\;\text{Hene: }\,\angle DEC \,=\,90\,+\,\frac{1}{2}\theta$

$\text{In }\Delta DEC:\:\angle EDC\,+\,\angle DEC\,+\,\angle ECD \:=\:180^o$
$\text{W\!e have: }\:\angle EDC\,+\,\left(90\,+\,\frac{1}{2}\theta\right)\,+\, \left(80\,-\,\frac{1}{2}\theta\right) \;=\;180$

$\text{Therefore: }\: \angle EDC \,=\,10^o$

 April 10th, 2015, 09:27 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,923 Thanks: 1123 Math Focus: Elementary mathematics and beyond $\displaystyle \angle{ABC}=\angle{ACB}=y$ $\displaystyle \angle{ADE}=\angle{AED}=x$ $\displaystyle 180-2x=180-20-2y=160-2y\implies x=10+y$ $\displaystyle \angle{DEC}=180-x=180-10-y,\angle{DCE}=y,\angle{EDC}=a$ $\displaystyle 180-10-y+y+a=180\implies a=10^{\circ}$ Thanks from aurel5 and Avlis Odraude

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