Question

Show that the sum of the three altitudes of a triangle is less than the sum of the three sides of the triangle .




For : @ Abhi ​

Answer 1

In ∆ABD,

∠D = 90°

and ∠B is acute

∠D > ∠B

∴ AB > AD ...(1)

| Side opposite to greater angle is longer

In ∆ABD,∠D = 90°and ∠B is acute∠D > ∠B∴ AB > A

In ∆ACD,

∠D = 90° and ∠C is acute.

∴ ∠D > ∠C

∴ AC > AD ...(2)

| Side opposite to greater angle is longer Adding (1) and (2), we have

AB + AC > 2AD ...(3)

Similarly, we can prove that,

BC + BA > 2BE ...(4)

| ∵ BE straight pi AC

and CA + CB > 2CF ...(5

| ∴ CF straight pi AB

Adding (3), (4) and (5), we get

⇒ 2(AB + BC + CA) > 2(AD + BE + CF)

⇒ AB + BC + CA > AD + BE + CF

⇒ AD + BE + CF < AB + BC + CA.