AD,BE and CF,the altitude of ∆ABC are equal .then (obtion A- AC =BC ,B-AD=AB,C-AB=CF]
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consider triangle ABE and triangle ACF
BE = CF (given)
angle A = angle A (common)
angle AEB = angle AFC = 90degrees
therefore by AAS congruency triangle ABE is congruent to triangle ACF
AB = AC (c.p.c.t.)
similarly, triangle DAC is congruent to triangle CBE
AC = BC (c.p.c.t.)
similarly, triangle BCF is congruent to triangle ABD
AB = BC (c.p.c.t.)
now,
AB = BC
AC = BC
AB = AC
this implies that
AB = BC = AC
therefore the triangle is equilateral
BE = CF (given)
angle A = angle A (common)
angle AEB = angle AFC = 90degrees
therefore by AAS congruency triangle ABE is congruent to triangle ACF
AB = AC (c.p.c.t.)
similarly, triangle DAC is congruent to triangle CBE
AC = BC (c.p.c.t.)
similarly, triangle BCF is congruent to triangle ABD
AB = BC (c.p.c.t.)
now,
AB = BC
AC = BC
AB = AC
this implies that
AB = BC = AC
therefore the triangle is equilateral
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