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Akshay_Budhkar
Posting a question wait lemme draw...
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prove that angle d = 90 + (1/2)( angle A)
ok,in triangle ABC, <ABC+<ACB+<BAC=180 or,<ABC+<ACB=180-<BAC <BAC=<A <CDB=<D in, DBC, <DBC+<BCD+<CDB=180 or, <ABC/2+<ACB/2+<CDB=180 or, (<ABC+<ACB)/2+<CDB=180 or, (180-<A)+<D=180 or, 90-<A/2+<D=180 or, <D=90+<A/2 ( proved )
sri used substitution to solve 2 eqs. we could also use elimination. Let y= angle CBD, and x= angle BCD 2y + 2x + A = 180 y + x + D = 180 multiply the first eq by -1/2 and add the 2 eqs then solve for D
angle<D=180-(<DBC+<DCB)=180-(<ABC+<ACB)/2=180-(180-<A)/2=180-90+<A/2=90+<A/2