Given: Parallel lines PQ and RS. Triangle ABD and BDC are such that
[tex]\begin{gathered} BD=CD \\ m\angle ABR=50\degree \\ m\angle ADB=65\degree \end{gathered}[/tex]
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,
[tex]\angle ADB=\angle DBC=65\degree[/tex]
Now since BD=CD, triangle BCD is an isosceles triangle. Hence,
[tex]\angle DBC=\angle DCB=65\degree[/tex]
Now, in triangle BCD, we have
[tex]\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}[/tex]
Now RS is a straight line. Hence at point B, we have
[tex]\begin{gathered} 50\degree+\angle ABD+\angle DBC=180\degree\text{ \lparen Linear pair\rparen} \\ \angle ABD=65\degree \end{gathered}[/tex]
Finally, in triangle ABD, we have
[tex]\begin{gathered} \angle A+\angle B+\angle D=180\degree \\ \angle A+65\degree+65\degree=180\degree \\ \angle A=50\degree \end{gathered}[/tex]
Now since in triangle ABD, we have
[tex]\angle ABD=\angle ADB[/tex]
The triangle ABD is isosceles.
Final Answer:
[tex]\begin{gathered} \angle BDC=50\degree \\ \angle ABD=65\degree \\ \angle BAD=50\degree \end{gathered}[/tex]
The triangle ABD is isosceles.
