Question

Prove that in an isosceles triangle any of its equal sides is greater than the straight line joining the vertex to any point on the base of the triangle.

Answer 1

Let the triangle be PQR.
PS QR, the straight line joining vertex P
to the line QR.
To prove : PQ > PT and PR > PT
In ΔPSQ,
PS2 + SQ² = PQ²         ....(Pythagoras theroem)
PS² = PQ² - SQ²          ....(i)
In ΔPST,
PS² + ST² = PT²           ....(Pythagoras theroem)
PQ² - SQ² = PT² - ST² ....(ii)
PQ - (ST + TQ)² = PT² - ST² ....[from (i) and (ii)]
PQ² - (ST² - 2ST x TQ + TQ²) = PT² - ST²
PQ² - (ST² - 2ST x TQ - TQ2 = PT² - ST
PQ² - PT² = TQ² + 2ST x TQ
PQ² - PT² = TQ x (2ST + TQ)
As, TQ x (2ST + TQ) > 0 always.
PQ² - PT² > 0
PQ² > PT²
PQ > PT
Also, PQ = PR
PR > PT.