Doubt by Suraj
Solution :
Given : AD, BE and CF are altitudes of ΔABC.
AD=BE=CF
To Prove : ABC is an equilateral triangle.
AD=BE=CF
To Prove : ABC is an equilateral triangle.
Proof :
Area of ΔABC
= ½×Base×Height
= ½×BC×AD — (1)
Again,
Area of ΔABC
= ½×Base×Height
= ½×AB×CF — (2)
Area of ΔABC
= ½×Base×Height
= ½×BC×AD — (1)
Again,
Area of ΔABC
= ½×Base×Height
= ½×AB×CF — (2)
From equation (1) and (2)
½×BC×AD=½×AB×CF
BC×AD=AB×CF
BC×AD=AB×AD [∵AD=CF]
BC=AB — (3)
½×BC×AD=½×AB×CF
BC×AD=AB×CF
BC×AD=AB×AD [∵AD=CF]
BC=AB — (3)
Similarly
BC=AC — (4)
From equation (3) and (4)
BC=AC — (4)
From equation (3) and (4)
AB=BC=CA
All sides of triangle ABC are equal.
So, ABC is an equilateral triangle.
Hence Proved.
Note : The above question can also be proved by using the concept of congruency. Can you do that by yourself?
Similar Question : In given figure the altitudes AD, BE and CF, the altitudes of triangle ABC are equal. Prove that ABC is an equilateral triangle.
Similar Question : The altitudes of ΔABC, AD, BE and CF are equal. Prove that ΔABC is an equilateral triangle.