Question

d is a point on the side BC of triangle ABC. show that perimeter triangle ABC>2AD

Answer 1

$\textbf{\huge{\underline{ Hello Mate! }}}$

To prove : Perimeter of triangle ABC > 2AD

Proof : ⤵⤵⤵

Since sum of two sides is great than third side.

In triangle ABD,

AB + BD > AD _____(1)

Since sum of two sides is great than third side

In triangle ACD,

AC + CD > AD ____(2)

Adding (1) and (2) we get,

AB + ( BD + CD ) + AC > AD + AD

AB + BC + AC > 2AD

Since perimeter of triangle is sum of all its side i.e. AB + BC + AC.

Perimter of tri ABC > 2AD

$\textbf{\large{ Q.E.D }}$

$\textbf{ Have great future ahead! }$

Answer 2

here is your answer OK dude



In triangle ABD,
AB + BD > AD [because, the sum of any
two sides of a triangle is
always greater than the
third side] ---- 1
In triangle ADC,
AC + DC > AD [because, the sum of any
two sides of a triangle is
always greater than the
third side] ---- 2
Adding 1 and 2 we get,
AB + BD + AC + DC > AD + AD
=> AB + (BD + DC) + AC > 2AD
=> AB + BC + AC > 2AD
Hence proved