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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.

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

Short Answer Type

307.

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

75 Views

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.

507 Views

309.
In the given Fig, PA, QB and RC are each perpendicular to AC. Prove that $1 over straight x plus 1 over straight z plus 1 over straight y.$

In ∆CBQ and ∆CAP
∠BCQ = ∠ACP [common]
and ∠CBQ = ∠CAP = 90°
Therefore, by AA similar condition
$increment CBQ space tilde space increment CAP$
$rightwards double arrow$ $space space CB over CA equals BQ over AP space space space space space space space space space space space space... left parenthesis straight i right parenthesis$

In ∆ABQ and ∆ACR
∠BAQ = ∠CAR [common]
and ∠ABQ = ∠ACR = 90°
Therefore, by AA similar condition
$increment ABQ tilde increment ACR$
$rightwards double arrow$ $AB over AC equals BQ over CR$ ...(ii)
Adding (i) and (ii), we get
$CB over CA plus AB over AC equals BQ over AP plus BQ over CR$

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$space space space space space space rightwards double arrow space space space space space space space space space space AC over AC equals straight y open parentheses 1 over straight x plus 1 over straight z close parentheses$
$rightwards double arrow space space space space space space space space space space space space space space space space 1 over straight y equals 1 over straight x plus 1 over straight z.$