# Please do for me no.79 n write steps also..

**To draw a parallelogram ABCD in which BC = 5 cm and AB = 3 cm and angle abc= 60 degree, follow the following steps:**(1) Draw a line segment AB = 3 cm

(2) At B, constrcut angle ABM = 60 degree.

(3) From BM, cut-off a line segment BC = 5 cm

(4) Now through the point A, draw AN||BC

(5) Through the point C, draw CD || BA, meeting AN at D.

Thus ABCD is the required parallelogram.

(6) Through diagonal BD, divide the parallelogram ABCD into two triangles BCD and ABD.

**To construct the triangle BD'C' similar to triangle BDC with scale factor 4:3, we have**(1) Below BD make an acute angle DBX.

(2) Along BX, mark four points B

_{1}

_{, }B

_{2}

_{,}B

_{3}

_{,}and B

_{4}such that BB

_{1}=B

_{1}B

_{2}=B

_{2}B

_{3}=B

_{3}B

_{4}

(3) Join B

_{3}D

(4) From B

_{4}, draw B

_{4}D' || B

_{3}D meeting BD produced at D'

(5) From D', draw C'D' || CD meeting BD produced at C'.

Thus, BC'D' is the required triangle whose sides are 4/3 times the corresponding side of triangle BCD.

Now, Draw thw line segment D'A' parallel to DA, where A' lies on the extended side BA.

**Justification:**Since AB || CD and CD || C'D'

Therefore, AB|| C'D' $\Rightarrow $AB' || C'D'

Similarly,

BC|| DA and DA |||D'A'

So, BC || D'A' $\Rightarrow $BC' || D'A'

Thus, BA'C'D' is a parallelogram.

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