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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.

Given:

increment FEC approximately equal to space increment GBD

and

space angle 1 space equals space angle 2

To Prove:

space space space space space space space increment ADE space tilde space increment ABC

Proof: In

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we have

space space space angle 1 space equals space angle 2

[given]

rightwards double arrow

AD = AE ...(i)
[sides opposite equal angles are equal]

space space space space increment GBD approximately equal to increment FEC

[given]

rightwards double arrow

BD = CE ...(ii) [by CPCT]
Dividing corresponding part of (i) by (ii), we get

AD over BD equals AE over EC

Therefore, by converse of Basic proportionality theorem,
DE || BC
Now, in ∆ADE and ∆ABC
∠ADE = ∠ABC
[corresponding angles]
∠AED = ∠ACB
[corresponding angles]
and ∠A = ∠A [common]
Therefore, by using AAA condition
∆ADE ~ ∆ABC.

1970 Views

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.

807 Views

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.

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