1. Show that in a right angled triangle, the hypotenuse is the longest side Let us consider ΔABC such that ∠B = 90º ∴ ∠A + ∠B + ∠C = 180º ∴ [∠A + ∠C] + ∠B = 180º ⇒ ∠A + ∠C = 90º ⇒ ∠A + ∠C= ∠B ∴ ∠B > ∠A and ∠B > ∠C ⇒ Side opposite to ∠B is longer than the side opposite to ∠A. i.e. AC > BC ...(1) Similarly, AC > AB ...(2) From (1) and (2), we get AC is the longest side. But AC is the hypotenuse of the triangle. Thus, hypotenuse is the longest side. <!--[if !supportLineBreakNewLine]--> <!--[endif]--> 2. In the adjoining figure, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB. ∠ABC + ∠PBC = 180º [Linear pair] and ∠ACB + ∠QCB = 180º [Linear pair] ∴ ∠ABC + ∠PBC = ∠ACB + ∠QCB But ∠PBC < ∠QCB [Given] ∴ ∠ABC > ∠ACB ⇒ [The side opposite to ∠ABC] > [The side opposite to ∠ACB] ⇒ AC > AB 3. In the figure, ∠B < ∠A and ∠C < ∠D. Show that AD > BC. ∵ ∠B< ∠A [Given] ⇒ ∠A> ∠B ∴ OB > OA [Side opposite to greater angle is longer] ...(1) Similarly, OC > OD ...(2) From (1) and (2), we have [OB + OC] > [OA + OD] ⇒ BC > AD or AD < BC 4. 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. Now, in ΔABC, AB < BC [∵ AB is the smallest side of quadrilateral ABCD] ⇒ BC > AB ∴ [Angle opposite to BC] < [Angle opposite to AB] ⇒ ∠BAC > ∠BCA ...(1) Again, in ΔACD, CD > AD [∵ CD is the longest side of the quadrilateral ABCD] ∴ [Angle opposite to CD] > [Angle opposite to AD] ⇒ ∠CAD > ∠ACD ...(2) Adding (1) and (2), we get [∠BAC + CAD] > [∠BCA + ∠ACD] ⇒ ∠A> ∠C Similarly, by joining BD, we have ∠B> ∠D 5. In the figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ. In ΔPQR, PS bisects ∠QPR [Given] ∴ ∠QPS = ∠RPS ∵ PR > PQ [Given] ∴ [Angle opposite to PR] > [Angle opposite to PQ] ⇒ ∠PQS > ∠PRS ⇒ [∠PQS + ∠QPS] > [∠PRS + ∠RPS] ...(1) [∵ ∠QPS = ∠RPS] ∵ Exterior ∠PSR = [∠PQS + ∠QPS] [∵ An exterior angle is equal to the sum of interior opposite angles] and Exterior ∠PSQ = [∠PRS + ∠RPS] Now, from (1), we have ∠PSR > ∠PSQ 6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest. Let us consider the ΔPMN such that ∠M = 90º Since, ∠M + ∠N + ∠P = 180º [Sum of angles of a triangle] ∵ ∠M = 90º [∵ PM ⊥ ℓ] ⇒ ∠N< ∠M ⇒ PM < PN ...(1) Similarly, PM < PN1 ...(2) PM < PN2 ...(3) From (1), (2) and (3), we have PM is the smallest line segment drawn from P on the line l. Thus, the perpendicular segment is the shortest line segment drawn on a line from a point not on it.