$In$ $triangle$ $ABC$ $bisectors$ $AA_1$, $BB_1$ $and$ $CC_1$ $are$ $drawn$. $Bisectors$ $AA_1$ $and$ $CC_1$ $intersect$ $segments$ $C_1B_1$ $and$ $B_1A_1$ $at$ $points$ $M$ $and$ $N$, $respectively$. $Prove$ $that$ $\angle$$MBB_1$ = $\angle$$NBB_1$.
Problem
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Tags: geometry
babu2001
20.03.2016 11:52
Here is a solution without using much calculation. Let d(O,XY) denote perpendicular distance of O from XY. Let AA' , BB' , CC' be the bisectors. It is well known that B'C' is the locus of points X satisfying d(X,BC) = d(X,AB) + d(X,AC) and AA' is the locus of points Y satisfying d(Y,AB) = d(Y,AC). Since M is intersection of B'C' and AA' , M satisfies both conditions. Hence, d(M,BC) = d(M,AB) + d(M,AC) = 2*d(M,AB) . Let perpendicular from M to AB and AC have feet M' and M" respectively. Then MM"/MM' = 2 (*). Similarly let the perpendicular from N to AC and to AB be N' and N" respectively. Then they will also satisfy NN"/NN' = 2 (**). From (*) and (**) we get MM"/MM' = NN"/NN' and angle M'MM" = N'NN" = 180* - A . Thus triangles M'MM" and N'NN" are similar so MM"M' = NN"N'. From cyclic quadrilaterals AM"MM' and AN"NN' we have MAM' = MM"M' and NAN' = NN"N'. Hence angles MAM' and NAN' are equal, which directly proves that angles MBB' and NBB' are equal