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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.

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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.

Given: In figure, PR > PQ and PS bisects ∠QPR.
To Prove: ∠PSR > ∠PSQ
Proof: In ∆PQR,
PR > PQ    | Given
∴ ∠PQR > ∠PRQ    ...(1)
| Angle opposite to longer side is greater
∵ PS is the bisector of ∠QPR
∴ ∠QPS = ∠RPS    ...(2)
In ∆PQS,
∠PQR + ∠QPS + ∠PSQ = 180° ...(3)
| ∵    The sum of the three angles of a A is 180°
In ∆PRS.
∠PRS + ∠SPR + ∠PSR = 180° ...(4)
| ∵    The sum of the three angles of a ∆ is 180°
From (3) and (4),
∠PQR + ∠QPS + ∠PSQ
= ∠PRS + ∠SPR + ∠PSR
⇒ ∠PQR + ∠PSQ = ∠PRS + ∠PSR
⇒ ∠PRS + ∠PSR = ∠PQR + ∠PSQ
⇒ ∠PRS + ∠PSR > ∠PRQ + ∠PSQ
| From (1)
⇒ ∠PRQ + ∠PSR > ∠PRS + ∠PSQ
| ∵ ∠PRQ = ∠PRS
⇒    ∠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.

1812 Views