Advertisements
Advertisements
Question
In the given figure,
AB || DC prove that
(i) ∆DMU ∼ ∆BMV
Solution
(i) Given `AB||DC`
In triangle DMU and BMV, we have
`∠ MUD =∠ MVB`
Each angle is equal to 90°
`∠ UMD =∠ VMB`
Each are vertically opposite angles.
Therefore, by AA-criterion of similarity
` ∆DMU ∼ ∆BMV`
APPEARS IN
RELATED QUESTIONS
In Figure below, if AB ⊥ BC, DC ⊥ BC and DE ⊥ AC, Prove that Δ CED ~ ABC.
In an isosceles ΔABC, the base AB is produced both the ways to P and Q such that AP × BQ = AC2. Prove that ΔAPC ~ ΔBCQ.
In each of the figures [(i)-(iv)] given below, a line segment is drawn parallel to one side of the triangle and the lengths of certain line-segment are marked. Find the value of x in each of the following :
In each of the following figures, you find who triangles. Indicate whether the triangles are similar. Give reasons in support of your answer.
ABCD is a trapezium having AB || DC. Prove that O, the point of intersection of diagonals, divides the two diagonals in the same ratio. Also prove that
In ∆ABC, AD and BE are altitude. Prove that\[\frac{ar\left( ∆ DEC \right)}{ar\left( ∆ ABC \right)} = \frac{{DC}^2}{{AC}^2}\]
In a quadrilateral ABCD, given that ∠A + ∠D = 90°. Prove that AC2 + BD2 = AD2 + BC2.
State SAS similarity criterion.
The areas of two similar triangles are 169 cm2 and 121 cm2 respectively. If the longest side of the larger triangle is 26 cm, what is the length of the longest side of the smaller triangle?
In the given figure, LM = LN = 46°. Express x in terms of a, b and c where a, b, c are lengths of LM, MN and NK respectively.
