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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.

2067 Views

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.

Given: ABCD is parallelogram in which BD is a diagonal, which intersects AE at F.

To Prove: DF.EF = FB.FA
Proof : In ∆AFD and ∆BFE
∠ADF = ∠EBF [alternate angles]
and ∠AFD = ∠BFE [vertically opposite angles]
Therefore, ∆DAF ~ ∆BEF
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581 Views

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.

1419 Views