Question

(c) In the diagram below:ARga nainP.50°B65%DNot drawn to scale(i) Calculate the angle BDC (ii) Calculate angle ABD (iii) Find angle BAD(iv) What type of triangle is triangle ABD ?CS

Answer 1

Given: Parallel lines PQ and RS. Triangle ABD and BDC are such that

$\begin{gathered} BD=CD \\ m\angle ABR=50\degree \\ m\angle ADB=65\degree \end{gathered}$

Required: To determine the triangle ABD type and calculate the angle BDC, ABD, and angle BAD.

Explanation: Since line PQ is parallel to line RS,

$\angle ADB=\angle DBC=65\degree$

Now since BD=CD, triangle BCD is an isosceles triangle. Hence,

$\angle DBC=\angle DCB=65\degree$

Now, in triangle BCD, we have

$\begin{gathered} \angle B+\angle C+\angle D=180\degree\text{ \lparen Angle sum property\rparen} \\ 65\degree+65\degree+\angle D=180\degree \\ \angle D=50\degree \end{gathered}$

Now RS is a straight line. Hence at point B, we have

$\begin{gathered} 50\degree+\angle ABD+\angle DBC=180\degree\text{ \lparen Linear pair\rparen} \\ \angle ABD=65\degree \end{gathered}$

Finally, in triangle ABD, we have

$\begin{gathered} \angle A+\angle B+\angle D=180\degree \\ \angle A+65\degree+65\degree=180\degree \\ \angle A=50\degree \end{gathered}$

Now since in triangle ABD, we have

$\angle ABD=\angle ADB$

The triangle ABD is isosceles.

Final Answer:

$\begin{gathered} \angle BDC=50\degree \\ \angle ABD=65\degree \\ \angle BAD=50\degree \end{gathered}$

The triangle ABD is isosceles.