Similar Triangles Questions

Solution : Since BD is a line and OC is a ray on it. \(\angle\) DOC + \(\angle\) BOC = 180 So,  \(\angle\) DOC + 125 = 180 \(\angle\) DOC = 55 In triangle CDO, we have : \(\angle\) CDO + \(\angle\) DOC + \(\angle\) DCO = 180 70 + 55 + \(\angle\) DCO = …

In the figure, \(\triangle\) ODC ~ \(\triangle\) OBA, \(\angle\) BOC = 125 and \(\angle\) CDO = 70. Find the \(\angle\) DOC, \(\angle\) DCO and \(\angle\) OAB. Read More »

Solution : (i)  In triangles ABC and PQR, we observe that \(\angle\) A = \(\angle\) B = 60, \(\angle\) P = \(\angle\) Q = 80 and \(\angle\) C = \(\angle\) R = 40 \(\therefore\)  By AAA criterion of similarity, \(\triangle\) ABC ~ \(\triangle\) PQR (ii)  In triangles ABC and PQR, we observe that \(AB\over QR\) …

State which pairs of triangle in the figure, 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 : Read More »

Solution : Given : A quadrilateral ABCD, in which the diagonals AC and BD intersect each other at O such that \(AO\over BO\) = \(CO\over DO\). To  Prove : ABCD is a trapezium. Construction : Draw EO || BA, meeting AD in E. Proof : In triangle ABD, EO || BA By basic proportionality theorem, …

The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(AO\over BO\) = \(CO\over DO\). Show that ABCD is a trapezium. Read More »

Solution : Given : A trapezium ABCD, in which the diagonals AC and BD intersect each other at O. To  Prove : \(AO\over BO\) = \(CO\over DO\) Construction : Draw EF || BA || CD, meeting AD in E. Proof : In triangle ABD, EF || AB By basic proportionality theorem, \(DO\over OB\) = \(DE\over …

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that \(AO\over BO\) = \(CO\over DO\). Read More »

Solution : Given : A triangle ABC in which D and E are mid point of sides AB and AC respectively. To Prove :  DE || BC Proof : Since sides AB and AC have D and E as mid points, \(\therefore\)   AD = DB  and  AE = EC \(\implies\)  \(AD\over DB\) = 1  and  …

Prove that the line joining the mid-points of two sides of a triangle is parallel to the third side. Read More »

Solution : Given : A triangle ABC in which D is the mid point of side AB and the line DE is drawn parallel to BC, meeting AC in E. To Prove :  AE = EC Proof : In triangle ABC, DE || BC By basic proportionality theorem, \(AD\over DB\) = \(AE\over EC\)      …

Prove that the line drawn from the mid-point of one side of triangle parallel to another side bisects the third side. Read More »

Solution : Given : O  is any point within triangle PQR, AB || PQ and AC || PR To Prove :  BC || QR Construction : Join BC Proof : In triangle OPQ, Given,      AB || PQ By basic proportionality theorem, \(OA\over AP\) = \(OB\over BQ\)                …

In the figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR. Read More »

Solution : In triangle PQO, Given,        DE || OQ By Basic proportionality theorem, we have \(PE\over EQ\) = \(PD\over DO\)           ……..(1) In triangle POR, Given,        DF ||  OR By Basic proportionality theorem, we have \(PD\over DO\) = \(PF\over FR\)           ……..(2) From …

In the figure, DE || OQ and DF || OR. Show that EF || QR. Read More »

Solution : In triangle BCA, Given,        DE || AC By Basic proportionality theorem, we have \(BE\over EC\) = \(BD\over DA\)           ……..(1) In triangle BEA, Given,        DF ||  AE By Basic proportionality theorem, we have \(BF\over FE\) = \(BD\over DA\)           ……..(2) From …

In the figure, if DE || AC and DF || AE, prove that \(BF\over FE\) = \(BE\over EC\). Read More »

Solution : In triangle ABC, Given,        LM || CB By Basic proportionality theorem, we have \(AM\over AB\) = \(AL\over AC\)           ……..(1) In triangle ACD, Given,        LN || CD By Basic proportionality theorem, we have \(AL\over AC\) = \(AN\over AD\)           ……..(2) From …

In the figure, if LM || CB and LN || CD, prove that \(AM\over AB\) = \(AN\over AD\). Read More »