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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.
Solution
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.
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