Question

The base angles of some isosceles triangle are 80°. Point D is on segment AC

and point E is on AB such that ∠DBC = 60 and ∠ECB 50 . Find the

angle EDB.

Answer 1

1. We know that $\angle$ ABC = $\angle$ ACB = $80^\circ$
2. Draw a line DH parallel to BC as shown and the length of DH being equal to the length of BC. Connect H to B. Then we have a parallelogram BCDH as depicted in the figure below.
3. Draw a PC with P connecting the line BD such that $\Delta$BCP becomes an equilateral triangle i.e. BC = BP = CP and $\angle$BCP = $\angle$BPC = $\angle$ CB = $60^\circ.$
4. $\angle$ DCP = $20^\circ$ because $\angle$ BCP = $60^\circ$ and $\angle$ BCD = $80^\circ$ . Also $\angle$ HBE = $20^\circ$. Hence, by side-angle-side, $\Delta$HBE = $\Delta$CPD.
5. Then, $\angle$ BHE = $\angle$ CDP = $\angle$ CDB = $40^\circ$ .
6. As $\angle$ BHD = $80^\circ$ , line HE becomes the angle bisector for $\angle$ BHD.
7. Also, BE is the angle bisector of $\angle$ DBH.
8. Therefore, the point E becomes the intersection of all three angles of the $\Delta$BHD also known as the incenter.
9. Then, ED is the bisector of $\angle$ BDH.
10. As $\angle$ BDH = $60^\circ$ , therefore, $\angle$ BDE = $30^\circ$ .