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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.

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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.

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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.

∠DOC + ∠BOC = 180° [Linear pair]
⇒ ∠DOC + 125° = 180
⇒ ∠DOC = 180 - 125
⇒ ∠DOC = 55°
In ∆DOC, we have
∠DOC + ∠ODC + ∠DCO = 180°
[Angle sum property of triangle]
⇒ 55° + 70° + ∠DCO = 180°
⇒ 125° + ∠DCO = 180°
⇒ ∠DCO =180° - 125°
⇒ ∠DCO = 55°
∵ ∆ODC ~ ∆OBA [Given]
∴ ∠OCD = ∠OAB
⇒ ∠DCO = ∠OAB
⇒ ∠OAB = ∠DCO
But ∠DCO = 55°
⇒ ∠OAB = 55°.