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Let F be the midpoint of AD. Then EF is an altitude of triangle ADE and is perpendictualr to AB.
Put triangle ABC on a coordinate system so point D is on the origin and side AB is on the x-axis. Let DF be the unit size of the system. Then point F is on (1,0), point A is on (2,0) and point B is on (-2,0). Let point E be at (1,t) and point C be at (p,q).
The measure of angle DAE (same as BAC) equals the measure of angle EDA since DE=EA in triangle DEA. The slope of DE is t, so then tan(EDA)=t. Thus, the measure of angle BAC equals arctan(t).
The slope of the line containing AE is (t-0)/(1-2) = -t. Then the equation of the line containng AE is (y-0) = -t(x-2). Since C, E, and A are colinear then C(p,q) is on the line. Substituting gives (q-0) = -t(p-2) which simplifies to q = -pt+2t.
Two equations can be formed from the distance formula for sides BC and CE. (-2-p)^2+(0-(-pt+2t))^2=2^2 and (p-1)^2+((-pt+2t)-t)^2=2^2. These equations simplify to (p-1)^2*(1+t^2)=4 and (p+2)^2+t^2*(p-2)^2=4.
Solving the first equation for t yields t = sqrt(4/(p-1)^2 - 1). Substituting the expression for t into the second gives (p-2)^2 + (p+2)^2 + 4*(p-2)^2/(p-1)^2 = 4. This equation simplifies to 2p^3-4p^2+3=0. The one real solution is p=-.739908.
With p=-.739908 then t=.566847 and angle BAC=29.5467 degrees. |
