13. In an equilateral triangle prove that any two altitudes on the sides from the vertices are of equal length
Answers
Answer:
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Step-by-step explanation:
Given : AD = BE = CF and angles A = B = C= 90°
To prove : Triangle ABC is equialteral.
Proof : In triangle ABE and ACF
Angle AEB = AFC ( 90° each )
BE = CF ( Given )
Angle A = A ( Common )
Hence Triangle ABE ~ ACF by AAS congruency
AB = AC ( c.p.c.t )
In triangle AOE and EOC
OE = OE ( Common )
Angle E = E ( 90° each )
AO = OC [ :. AD = FC, their halfs OA = OC ]
Hence triangle AOE ~ EOC by RHS congruency
AE = EC ( c.p.c.t )
In triangle ABE and BCE
Angle E = E ( 90° each )
BE = BE ( common )
AE = EC ( proved above )
Hence, triangle ABE ~ BCE by SAS congruency.
AB = BC ( c.p.c.t )
As AB = AC and AB = BC, so AC = BC.
Hence, AB = BC = CA.
HENCE PROVED.
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Given : AD = BE = CF and angles A = B = C= 90°
To prove : Triangle ABC is equialteral.
Proof : In triangle ABE and ACF
Angle AEB = AFC ( 90° each )
BE = CF ( Given )
Angle A = A ( Common )
Hence Triangle ABE ~ ACF by AAS congruency
AB = AC ( c.p.c.t )
In triangle AOE and EOC
OE = OE ( Common )
Angle E = E ( 90° each )
AO = OC [ :. AD = FC, their halfs OA = OC ]
Hence triangle AOE ~ EOC by RHS congruency
AE = EC ( c.p.c.t )
In triangle ABE and BCE
Angle E = E ( 90° each )
BE = BE ( common )
AE = EC ( proved above )
Hence, triangle ABE ~ BCE by SAS congruency.
AB = BC ( c.p.c.t )
As AB = AC and AB = BC, so AC = BC.
Hence, AB = BC = CA.
HENCE PROVED.