Let $ABC$ be an isosceles triangle with $AB = BC$. Point $E$ lies on the side $AB$, and $ED$ is the perpendicular from $E$ to $BC$. It is known that $AE = DE$. Find $\angle DAC$.
Problem
Source: Sharygin First Round 2013, Problem 1
Tags: geometry proposed, geometry
09.04.2013 02:05
The answer is $\boxed{45^\circ}$. Let $\angle BAD = \alpha$ and $\angle DAC = \beta$. Since triangle $AED$ is isosceles, $\angle ADE = \angle EAD = \alpha$ and thus $\angle CDA = 90^\circ - \alpha$. Then $\angle ACB = 90^\circ + \alpha - \beta$, but since $\angle ACB = \angle BAC$, it follows that \[90^\circ + \alpha - \beta = \alpha + \beta\] and so it follows that $\beta = \angle DAC = 45^\circ$.
19.07.2013 08:15
let $\angle ABC=x$ . then $\angle BED=\pi/2-x$ . as $AE=ED$ , so , $\angle EDA=\pi/4-x/2$. so , $\angle ADC=\pi/4+x/2$ . also , $\angle BCA=\pi-x/2$ . so , $\angle DAC=\pi/4$
19.07.2013 22:09
Obviously $BC$ is tangent to the circle $C(E,EA)$, and let this circle intersect 2nd time $AC$ at $F$. $AF=AE\implies EF\parallel BC\iff EF\bot ED$, hence $\angle DEF=90^\circ$ and $\angle DAF$ is half of $\angle DEF$, done. Best regards, sunken rock
10.08.2013 08:10
Suppose ${\angle}ABC=k$,so ${\angle}BED=90^{\circ} -k$,${\angle}DEA=90^{\circ} +k$,hence ${\angle}EAD=45^{\circ} -k/2$ again ${\angle}BAC=90^{\circ} -k/2$,so ${\angle}DAC={\angle}BAC-{\angle}BAD=45^{\circ}$
27.01.2015 13:52
LET EAD=EDA =Y LET BAC=BCA=X BED+EBD=90 2Y+(180-2X)=90 X-Y=45 DAC=BAC-BAD=X-Y=45
24.05.2015 21:29
robinpark wrote: Let $ABC$ be an isosceles triangle with $AB = BC$. Point $E$ lies on the side $AB$, and $ED$ is the perpendicular from $E$ to $BC$. It is known that $AE = DE$. Find $\angle DAC$. Let $\angle DAE \equiv \angle EDA \equiv x$ $\implies$ $\angle BED \equiv \angle EAD + \angle EDA = 2x$ $\implies$ $\angle EBD \equiv 90 - 2x$ $\implies$ $\angle DAC \equiv \angle BAC - x \equiv \frac{180 - 90 + 2x}{2} - x \equiv 45$.
22.01.2016 17:49
What does $ED$ is the perpendicular from $E$ mean as $E$ is on $AB$?
22.01.2016 21:05
robinpark wrote: Let $ABC$ be an isosceles triangle with $AB = BC$. Point $E$ lies on the side $AB$, and $ED$ is the perpendicular from $E$ to $BC$. It is known that $AE = DE$. Find $\angle DAC$. $\angle EAD= \angle EDA$ , $\angle ADC=90-\angle EAD$ , $180= \angle EAD + 2\angle DAC + 90 - \angle EAD$ , $\angle DAC =45$
27.09.2023 16:43
Since AB=BC SO, angle BAC=angle ACB=x angle ABC=180°-2x angle EDB=90° So, angle DEB=2x-90° angle DEA=180°-(2x-90°)=270°-2x In ∆ AED, AE=DE angle EAD=angle EDA angle DEA=270°-2x So, angle EAD=x-45° angle BAC=x So, angle DAC=45°