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Triangles

Long Answer Type

321. Prove that three times the sum of the squares of the sides of a triangle is equal to four times the sum of the squares of the medians of the triangle.

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322. Points P, Q, R and S in that order are dividing a line segment joining A(2, 6) and B(7, 4) in five equal parts. Find the co-ordinates of P and R.

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323. BL and CM are median of a triangle ABC right angled at A. Prove that 4(BL² + CM²) = 5BC².

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324.

In the given Fig, ABC is a right triangle, right angled at B. AD and CE are two medians drawn from A and C respectively. If AC = 5 cm and $AD equals fraction numerator 3 square root of 5 over denominator 2 end fraction cm comma$ find the length of CE.

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Short Answer Type

325. In an isosceles triangle ABC, with AB = AC, BD is perpendicular from B to the side AC. Prove that BD² - CD² = 2CD.AD.

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Long Answer Type

326. ABC is a triangle in which AB = AC and D is any point in BC. Prove that AB² - AD² = BD.CD.

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327. In a right triangle ABC, right angled at C.P. and Q are the points on the sides CA and CB respectively, which divdes these sides in the ratio 2:1. Prove that
(i) 9AQ² 9AC² + 4BC²(ii) 9BP² 9BC² + 4AC²(iii) 9(AQ² + BP²) = 13AB².

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328. In the given figure ABC is a triangle right angled at B. P and Q trisect BC. Prove that: 8AP² = 3AC² + 5AQ²

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329.

In the given Fig., ∠ACB = 90° and CD ⊥ AB. Prove that $BC squared over AC squared equals BD over AD.$

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330. P and Q are points on sides CA and CB respectively of ∆ABC, right angled at C. Prove that AQ² + BP² = AB² + PQ².

Given: A right ∆ABC right angled at C.

To prove:
AQ² + BP² = AB² + PQ²Proof: In right ∆ACQ, we have
AQ² = AC² + CQ²    ...(i)
[Using Pythagoras theorem]
In right ∆PCB, we have
BP² = PC² + BC²    ...(ii)
[Using Pythagoras theorem]
Adding (i) and (ii), we get
AQ² + BP² = (AC² + BC²) + (PC² + CQ²)
⇒ AQ² + BP² = AB² + PQ²    Proved.