Advertisements
Advertisements
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.
Advertisements
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.
RELATED QUESTIONS
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
In a triangle PQR; QR = PR and ∠P = 36o. Which is the largest side of the triangle?
D is a point in side BC of triangle ABC. If AD > AC, show that AB > AC.
In the following figure ; AC = CD; ∠BAD = 110o and ∠ACB = 74o.
Prove that: BC > CD.
“Caste inequalities are still prevalent in India.” Examine the statement.
In ΔABC, the exterior ∠PBC > exterior ∠QCB. Prove that AB > AC.
ABCD is a quadrilateral in which the diagonals AC and BD intersect at O. Prove that AB + BC + CD + AD < 2(AC + BC).
ABCD is a trapezium. Prove that:
CD + DA + AB > BC.
In ΔPQR, PS ⊥ QR ; prove that: PQ + PR > QR and PQ + QR >2PS.
In the given figure, T is a point on the side PR of an equilateral triangle PQR. Show that PT < QT
