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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.

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Long Answer Type

308. Through (the vertex D of a parallelogram ABCD, a line is drawn to intersect the sides BA and BC produced at E and F respectively. Prove that $DA over AE equals FB over BE equals FC over CD.$

In ∆DCF and ∆EAD
∠2 = ∠4 [corresponding angles]
and ∠1 = ∠3 [corresponding angles]
Therefore, by using AA similar condition
 $increment DCF tilde increment EAD$

$rightwards double arrow$ $space DC over EA equals CF over AD equals DF over ED$

$rightwards double arrow$ $DC over EA equals CF over AD$

$rightwards double arrow$ $DC over CF equals AE over AD$

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[Taking reciprocals of both sides]
Now, in ∆EAD and ∆EBF
∠4 = ∠4    [common]
and    ∠3 = ∠1 [corresponding angles]
Therefore, by using AA similar condition
$increment EAD space tilde space increment EBF$

$rightwards double arrow$ $EA over EB equals AD over BF equals ED over EF$

$rightwards double arrow$    $EA over EB equals AD over BF$

$rightwards double arrow$ $AD over AE equals BF over BE space space space space space space space space... left parenthesis ii right parenthesis$
Comparing (i) and (ii), we get
$DA over AE equals FB over BE equals FC over CD.$