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Triangles

Long Answer Type

181. In the adjacent figure, if LM || CB and LN || CD, prove that

AM over AB equals AN over AD

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182. In the given figure. DE || AC and DF || AE. Prove that

BF over FE equals BE over EC.

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183. In the given Fig. DE || OQ and DF || OR. Show that EF || QR.

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184. In the given fig, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

220 Views

Short Answer Type

185. Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

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186. Using Theorem 6.2 (N.C.E.R.T.), prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

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Long Answer Type

187. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that $AO over BO equals CO over DO.$

Given ABCD is a trapezium in which AB ||DC and diagonals AC and BD intersect each other at O.
To prove: $space AO over BO equals CO over DO$

Construction: Draw EO || DC || AB
In ∆ADC, we have

EO parallel to DC

rightwards double arrow space space space space space space space space space AO over OC equals AE over ED space space space space space space space space space space space... left parenthesis straight i right parenthesis

[by converse of Basic proportionality theorem]
In ∆ADB, we have

EO parallel to DC

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[by converse of Basic proportionality theorem]
In ∆ADB, we have

space space EO parallel to AB

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[Taking reciprocals of both side]
Comparing (i) and (ii), we get

AO over OC equals OB over OD

rightwards double arrow

AO over BO equals CO over DO space Hence space Proved.

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188. The diagonals of a quadrilateral ABCD intersect each other at point O such that

AO over BO equals CO over DO.

Show that ABCD is a trapezium.

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189. State which pairs of triangles in the given Fig, are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

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Short Answer Type

190.

In the given fig. ∆ODC ~ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

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