1. Two line segments AB and AC include an angle of 60⁰, where AB=5cm and AC=7cm. locate points P and Q on AB and AC, respectively such that AP=3/4AB and AQ=1/4ac. join Pand Q and measure the length PQ.
2. Draw an isosceles triangle ABC in which AB=AC=6cm and BC=5cm. construct a triangle PQR similar to triangle ABC in which PQ=8cm. also, justify the construction.
3. Draw a parallelogram ABCD in which BC =5cm, AB=3cm and angle ABC=60⁰, divide it into triangles BCD and ABD by the diagonal BD. construct the triangle BD'C' similar to triangle BDC with scale factor 4/3. Draw the line segment D'A' parallel to DA where A' lies on extended side BA. Is A'BC'D' a parallelogram?

$1^{st}$ PART

Here in the given figure $\angle$BAC = 60º
And $$\begin{gathered} AP = \frac{3}{4}\times AB = \frac{3}{4}\times 5 cm = \frac{15}{4}cm \\ and AQ= \frac{1}{4}\times AC = \frac{1}{4}\times 7 cm = \frac{7}{4}cm \end{gathered}$$
So here two sides and an angle between them is given so using the cosine rule we can get the measure of third side  $$\begin{gathered} c^2=a^2+b^2- 2 ab \cos c \\ so P{Q}^2=A{P}^2+A{Q}^2-2\times a\times b\times \cos c \\ \Rightarrow P{Q}^2=(\frac{15}{4}{)}^2+(\frac{7}{4}{)}^2-2\times \frac{15}{4}\times \frac{7}{4}\times COS 60° \\ \Rightarrow P{Q}^2=\frac{169}{16} \\ \Rightarrow PQ = \frac{13}{4}cm = 3.25 cm \end{gathered}$$

Now for the $2^{nd}$ part
As the two triangles are similar then the sides are in proportion
So $\frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}$
 AB , AC and PQ is given so $PR=\frac{\left(PQ\right)\times \left(AC\right)}{AB}=\frac{8\times 6}{6}=8cm$

Similarly $QR=\frac{(PQ\times BC)}{AB}=\frac{(8\times 5)}{6}=\frac{40}{6}cm$


So for constructing $∆$ABC , first draw BC = 5 cm and then from B and C draw arcs so that all points on the two arcs are at a distance AB=6 cm and AC =6cm  from B and C respectively. So the point where these two arcs intersect each other that will be the vertex A.​


Steps for the construction of $∆$PQR.
Extend B to Q and C to R such that BQ=  $\frac{5}{6}$cm and CR =$\frac{5}{6}$ cm
Draw lines QP $\parallel$BA and RP$\parallel$CA  such that QP = RP = 6cm.
​QP and RP will meet at P.

Justification for the construction:-
As while constructing $∆$PQR, we just draw lines which are parallel to two sides of the $∆$ABC

so $$\begin{gathered} \angle CBA = \angle RQP (correspnding angles as BA\parallel QP) \\ \angle BCA = \angle QRP (corresponding angles as CA\parallel RP) \\ \angle BAC = \angle QPR (\sin ce (180°-(\angle CBA+\angle BCA))= (180°-(\angle RQP+\angle QRP)\left)\right) \\ So by AAA similarity ,∆PQR~∆ABC \end{gathered}$$


In the $3^{rd}$ part

yes it is absolutely true that A'BC'D' will be a parallelogram as while drawing similar triangles ,we keep in mind that the sides are increased in proportion
Now the simple reason is that here  all sides are  are increased in same proportion i.e the scale factor is $\frac{4}{3}$
BC is increased to BC' in same proportion as BD to BD' and also CD to CD'
So  BA is increased to BA' in same proportion 
and one important observation is that the diagonal BD remains the diagonal only the length is increased .
Opposite sides are also parallel.

So in short one can say that we are here stretching the point D to a point D' of the parallelogram diagonally keeping point B fixed...