Show that of all line segments drawn from a given point not on i

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 Multiple Choice QuestionsShort Answer Type

491. ABCD is a quadrilateral in which AB = BC and AD = CD. Show that BD bisects both the angles ABC and ADC.  
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492. Show that in a right angled triangle, the hypotenuse is the longest side.
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493. In figure, sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC <  ∠QCB. Show that AC >  AB.


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494. In figure, ∠B <  ∠A and ∠C <  ∠D. Show that AD <  BC.

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495. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠A >  ∠C and ∠B >  ∠D.


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496. In figure, PR >  PQ and PS bisects ∠QPR. Prove that ∠PSR >  ∠PSQ.



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497. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.


Given: l is a line and P is a point not lying on l. PM ⊥ l. N is any point on l other than M.


Given: l is a line and P is a point not lying on l. PM ⊥ l. N is an

To Prove: PM < PN.
Proof: In ∆PMN,
∠M = 90°
∴ ∠N is an acute angle.
| Angle sum property of a triangle
∴ ∠M >  ∠N
∴ PN >  PM
| Side opposite to greater angle is greater
⇒ PM <  PN.

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498. Prove that the sum of three altitudes of a triangle is less than the sum of the three sides of the triangle.
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499. In figure, AD is a median of ∆ABC. Prove that AB + AC >  2AD.

[Hint. Produce AD to E such that AD = DE and join C and E.]

OR

Prove that the sum of any two sides of a triangle is greater than twice the length of median drawn to the third side. 

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500.

Prove that the sum of the three sides of a triangle is greater than the sum of its three medians.

OR

Prove that the perimeter of a triangle is greater than the sum of its three medians.

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