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Show that in a right angled triangle, the
hypotenuse is the longest side.
Let us consider a right-angled triangle ABC, right-angled at B.
In ΔABC,
∠A + ∠B + ∠C = 180° (Angle sum property of a triangle)
∠A + 90º + ∠C = 180°
∠A + ∠C = 90°
Hence, the other two angles have to be acute (i.e., less than 90º).
∴ ∠B is the largest angle in ΔABC.
⇒ ∠B > ∠A and ∠B > ∠C
⇒ AC > BC and AC > AB
[In any triangle, the side opposite to the larger (greater) angle is longer.]
Therefore, AC is the largest side in ΔABC.
However, AC is the hypotenuse of ΔABC. Therefore, hypotenuse is the longest side in a right-angled triangle.
In Figure, ∠ B < ∠ A and ∠ C < ∠ D. Show that AD < BC.
In ΔAOB,
⇒ AO < BO (Side opposite to smaller angle is smaller) ... (1)
In ΔCOD,
∠C < ∠D
⇒ OD < OC (Side opposite to smaller angle is smaller) ... (2)
On adding equations (1) and (2), we obtain
AO + OD < BO + OC
AD < BC
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Figure). Show that ∠ A > ∠ C and ∠ B > ∠ D.
Let us join AC.
In ΔABC,
AB < BC (AB is the smallest side of quadrilateral ABCD)
∴ ∠2 < ∠1 (Angle opposite to the smaller side is smaller) ... (1)
In ΔADC,
AD < CD (CD is the largest side of quadrilateral ABCD)
∴ ∠4 < ∠3 (Angle opposite to the smaller side is smaller) ... (2)
On adding equations (1) and (2), we obtain
∠2 + ∠4 < ∠1 + ∠3
⇒ ∠C < ∠A
⇒ ∠A > ∠C
Let us join BD.
In ΔABD,
AB < AD (AB is the smallest side of quadrilateral ABCD)
∴ ∠8 < ∠5 (Angle opposite to the smaller side is smaller) ... (3)
In ΔBDC,
BC < CD (CD is the largest side of quadrilateral ABCD)
∴ ∠7 < ∠6 (Angle opposite to the smaller side is smaller) ... (4)
On adding equations (3) and (4), we obtain
∠8 + ∠7 < ∠5 + ∠6
⇒ ∠D < ∠B
⇒ ∠B > ∠D
In Figure, PR > PQ and PS bisects ∠ QPR. Prove that ∠ PSR > ∠ PSQ.
As PR > PQ,
∴ ∠PQR > ∠PRQ (Angle opposite to larger side is larger) ... (1)
PS is the bisector of ∠QPR.
∴∠QPS = ∠RPS ... (2)
∠PSR is the exterior angle of ΔPQS.
∴ ∠PSR = ∠PQR + ∠QPS ... (3)
∠PSQ is the exterior angle of ΔPRS.
∴ ∠PSQ = ∠PRQ + ∠RPS ... (4)
Adding equations (1) and (2), we obtain
∠PQR + ∠QPS > ∠PRQ + ∠RPS
⇒ ∠PSR > ∠PSQ [Using the values of equations (3) and (4)]
Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Let us take a line l and from point P (i.e., not on line l), draw two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle.
In ΔPNM,
∠N = 90º
∠P + ∠N + ∠M = 180º (Angle sum property of a triangle)
∠P + ∠M = 90º
Clearly, ∠M is an acute angle.
∴ ∠M < ∠N
⇒ PN < PM (Side opposite to the smaller angle is smaller)
Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in comparison to them.
Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
In Figure, sides AB and AC of triangle ABC are extended to points P and Q respectively. Also, ∠ PBC < ∠ QCB. Show that AC > AB.
In the given figure,
∠ABC + ∠PBC = 180° (Linear pair)
⇒ ∠ABC = 180° − ∠PBC ... (1)
Also,
∠ACB + ∠QCB = 180°
∠ACB = 180° − ∠QCB … (2)
As ∠PBC < ∠QCB,
⇒ 180º − ∠PBC > 180º − ∠QCB
⇒ ∠ABC > ∠ACB [From equations (1) and (2)]
⇒ AC > AB (Side opposite to the larger angle is larger.)
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