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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.
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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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