Question

if s is the circumcenter of a triangle abc and d,e,f are the feet of the altitudes of triangle abc then prove that sb is perpendicular to df

Answer 1

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Define K as midpoint of BC. Call the intersection of BS and DF as P. Since S is the center of the circle of ABC, we have ∠BSC=2∠A. Since SB=SC, SK bisects ∠BDF therefore ∠BSK=∠A. On the other hand since D and F are feet of the altitudes CAFD is cyclic, hence ∠BDF=∠A. Now ∠A=∠BSK=∠BDF tells us KSPD is a cyclic quadrilateral. Hence ∠SKB=∠DPS=90∘.

Answer 2

${\huge{\underline{\underline{\mathcal {\red{♡Question♡}}}}}}$

If s is the circumcenter of a triangle abc and d,e,f are the feet of the altitudes of triangle abc then prove that sb is perpendicular to DF.

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${\huge{\underline{\underline{\mathcal {\red{♡Solution♡}}}}}}$

Define K as midpoint of BC.

Call the intersection of BS and DF as P.

Since S is the center of the circle of ABC,

we have,

• ∠BSC = 2∠A.

• Since SB = SC,

• SK bisects ∠BDF

therefore,

• ∠BSK = ∠A.

On the other hand since D and F are feet of the altitudes CAFD is cyclic.

hence,

• ∠BDF=∠A.

Now ,

• ∠A = ∠BSK = ∠BDF.........tells us KSPD is a cyclic quadrilateral.

Hence ∠SKB = ∠DPS = 90∘.

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