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Triangles

Short 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.

Given: In figure, ∠B < ∠A and ∠C < ∠D.
To Prove: AD < BC

Proof: ∠B < ∠A    | Given
∴ ∠A > ∠B
∴ OB > OA    ...(1)
| Side opposite to greater angle is longer
∠C < ∠D    | Given
∴ ∠D > ∠C
∴ OC > OD    ...(2)
| Side opposite to greater angle is longer
From (1) and (2), we get
OB + OC > OA + OD
⇒ BC > AD
⇒ 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.

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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.]

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.

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

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