document.write( "Question 252016: Q.2:- Let ABC be a triangle in which AB=AC and let I be its in-centre. suppose BC=AB+AI.find the angle BAC? \n" ); document.write( "
Algebra.Com's Answer #484268 by plastery(3)\"\" \"About 
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The angle BAC that satisfies the relation BC=AB+AI is the right angle.
\n" ); document.write( "To get to this solution with some trigonometry, let's call x the angle BAC and l the lenght of AB..
\n" ); document.write( "As the triangle ABC is isosceles and the sum of the internal angles of any triangle is 2 right angles, then the angle \"ABC=ACB=%28PI-x%29%2F2=PI%2F2+-X%2F2\".
\n" ); document.write( "Let me remind that the Incenter of a triangle can be found as the intersection of any two internal angle bisectors.
\n" ); document.write( "Let's call H the intersection of the bisector of BAC and BC.
\n" ); document.write( "AH is also the altitude relative to the base BC. So ABH is a right triangle (right in H).
\n" ); document.write( "For the definition of sine \"BH=BA%2Asin%28x%2F2%29=l+%2Asin%28x%2F2%29\"
\n" ); document.write( "For the Pythagorean theorem
\n" ); document.write( "The angle IBH is half of ABH that is \"IBH=ABC+%2F+2=+PI%2F4-x%2F4\"
\n" ); document.write( "For the definition of tangent \"IH=BH%2Atan%28IBH%29=l%2Asin%28x%2F2%29%2Atan%28PI%2F4-x%2F4%29\"
\n" ); document.write( "Finally \"AI=AH-IH=l%2Acos%28x%2F2%29-l%2Asin%28x%2F2%29%2Atan%28PI%2F4-x%2F4%29\"
\n" ); document.write( "Therefore, as \"BC=2%2ABH\", the relation \"BC=AB%2BAI\" can be written \"2%2Al+%2Asin%28x%2F2%29=l%2Bl%2Acos%28x%2F2%29-l%2Asin%28x%2F2%29%2Atan%28PI%2F4-x%2F4%29\" and with some trivial passage
\n" ); document.write( "\"2%2Asin%28x%2F2%29=%281%2Bcos%28x%2F2%29-sin%28x%2F2%29%2Atan%28PI%2F4-x%2F4%29%29\"
\n" ); document.write( "Still Algebra:
\n" ); document.write( "\"sin%28x%2F2%29%2A%282%2Btan%28PI%2F4-x%2F4%29%29=1%2Bcos%28x%2F2%29\"
\n" ); document.write( "Now we apply the sum identity of the tangent to obtain:
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\n" ); document.write( "Known that \"tan%28PI%2F4%29=1\", we obtain:
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\n" ); document.write( "Now let's apply the half angle formula for the tangent \"tan%28alpha%2F2%29=%281-cos%28alpha%29%29%2Fsin%28alpha%29\"
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\n" ); document.write( "Again Algebra and provided that \"sin%28x%2F2%29%3C%3E0\" i.e \"x%3C%3E0\" we obtain:
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\n" ); document.write( "then
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\n" ); document.write( "and
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\n" ); document.write( "and provided that \"%281%2Bsin%28x%2F2%29-cos%28x%2F2%29%29%3C%3E0\"
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\n" ); document.write( "and
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\n" ); document.write( "and
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\n" ); document.write( "and
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\n" ); document.write( "and
\n" ); document.write( "\"2%2Asin%5E2%28x%2F2%29-2%2Asin%28x%2F2%29%2Acos%28x%2F2%29%29=0\"
\n" ); document.write( "and
\n" ); document.write( "\"2%2Asin%28x%2F2%29%2A%28sin%28x%2F2%29-cos%28x%2F2%29%29=0\"
\n" ); document.write( "The first term cannot be zero for the previous condition, so the possible solution must satisfy:
\n" ); document.write( "\"sin%28x%2F2%29-cos%28x%2F2%29=0\"
\n" ); document.write( "that happens when \"x%2F2=PI%2F4\" i.e. \"x=PI%2F2\".
\n" ); document.write( "The general solution would be \"x=2%2API%2An-3%2F2%2API\" with n any integer, but for the geometric nature of the problem we can rest on \"n=1\".\r
\n" ); document.write( "\n" ); document.write( "Eventually we check the solution that is that in case of a right/isosceles triangle the relation BC=AB+AI is satisfied.
\n" ); document.write( "With the same notation above:
\n" ); document.write( "\"BC=sqrt%282%2Al%5E2%29=l%2Asqrt%282%29=l%2A%281%2B%28sqrt%282%29-1%29%29\"
\n" ); document.write( "BH is such that \"l%5E2=2%2ABH%5E2\" therefore \"BH=l%2Fsqrt%282%29\"
\n" ); document.write( "As \"ABC=PI%2F2\" then \"ABH=PI%2F4\" and \"IBH=PI%2F8\", then
\n" ); document.write( "But \"AH=BH=l%2Fsqrt%282%29\" and
\n" ); document.write( "So \"AB%2BAI=l%2Bl%2A%28sqrt%282%29-1%29=l%2Asqrt%282%29\" and \"BC=2%2ABH=2%2Al%2Fsqrt%282%29=l%2Asqrt%282%29\"
\n" ); document.write( "Quod Demonstrandum Erat
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