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Triangles

Long Answer Type

301. In ∆ABC, let P and Q he points on AB and AC respectively such that PQ || BC. Prove that the median AD bisects PQ.

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Short Answer Type

302. ABC is a triangle in which AB = AC and D is a point on AC such that BC² = AC × CD. Prove that BD = BC.

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Long Answer Type

303.

In the given Fig, ∠1 = ∠2. Prove that ∆ADE ~ ∆ABC.

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304.

In the given Fig. DEFG is a square and ∠BAC = 90°. Prove that
(i) ∆AGF ~ ∆DBG.
(ii) ∆AGF ~ ∆EFC.
(iii) ∆DBG ~ ∆AEFC
(iv) DE² = BD × EC.

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Short Answer Type

305. The diagonal BD of a parallelogram ABCD intersects AE at a point F, where E is any point on the side BC. Prove DF.EF = FB.FA.

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Long Answer Type

306. Prove that the line segments joining the mid points of the sides of a triangle form four triangles, each of which is similar to the original triangle.

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Short Answer Type

307.
In the given Fig., AC ⊥ BC, BD ⊥ BC and DE ⊥ AB. Prove that ABC ~ BDE.

Given: AC ⊥ BC, BD ⊥ BC and DE ⊥ AB.
To Prove : ∆ABC ~ ∆BDE
Proof: ∵ ∠1 = 90°    [given]
⇒    ∠2 + ∠3 = 90°    ...(i)
and    ∠3 + ∠4 = 90° ...(i) [given]
Comparing (i) and (ii), we get
⇒    ∠2 + ∠3 = ∠3 + ∠4
⇒    ∠2 = ∠4
or ∠BAC = ∠DBE
Now, in ∆ABC and ∆BDE
∠BAC = ∠DBE [alternate angles]
and ∠ACB = ∠BED = 90°
Therefore, by using AA similar condition
∆ABC ~ ∆BDE.

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