In the figure ADB CBD and ABD CDB AB 3x 4 BC x 4y CD

In the figure, ADB = CBD and ABD = CDB. AB = 3x + 4, BC = x - 4y, CD = 5x + y and AD = 3x - y. (a) How are the triangles related? (b) Find x, y, AB and BC.

Solution

In the given figure, ADB and the CBD are alternate angles. Similarly, the ABD and the CDB are alternate angles. Further, since the alternate angles are equal, hence AB is parallel to DC and AD is parallel to BC. Therefore, the given figure is a parallelogram. Now, since the opposite sides of a parallelogram are equal, AB = DC. Similarly, AD = BC.

(a). In the s ABD and BCD, the side BD is common, AB = DC and and AD = BC. Hence the s ABD and BCD are congruent.

(b). Since AB =DC , hence 3x +4 = 5x+y or, y = 3x-5x+4 or, y = -2x+4…(1).Similarly, AD = BC, so that 3x-y = x-4y or, 4y –y = x-3x or, 3y = -2x or, y = -(2/3)x…(2).

From the equations, 1 and 2, we have -2x+4 = y = -2x/3 so that 2x-2x/3 = 4 or, 4x/3 = 4. Hence x = 4*3/4 or, x = 3.Then , on substituting x = 3 in the 1^st equation, we have y = -2*3+4 = -6+4 or, y = -2. Then AB = 3x+4 = 3*3+4 = 9+4 or, AB= 13. Also,BC = x-4y = 3-4*(-2) = 3+8 or, BC = 11.