Question

in triangle abc, d is the midpoint of ac such that cd = bd. show that angle abc = 90°

Answer 1

Construction :  Draw line passing through points B and D

                         Extend BD to DE such that BD = DE

                        Draw line passing through points E and C

Solution : In ΔADB and ΔEDC

                  DB = DE

                ∠BDA = ∠CDE

               AD = DE

            ∴ ΔADB ≅ ΔEDC  (SAS Congruency)

               ∠ABD = ∠CED ( Corresponding parts of a congruent triangle )

              ∴AB║EB (∵ ∠ABD = ∠CED and both angles are alternate interior                               angles)

                 If  ΔADB ≅ ΔEDC

              Then   ΔADB  + Δ BDC ≅ ΔEDC + Δ BDC

            ∴ΔABC ≅ ΔECB

                ∠ABC = ∠ ECB

              Also ∠ABC + ∠ ECB = 180

                2∠ABC = 180

              ∠ABC = 180

Answer 2

In ΔADB, AD = BD

∠DAB = ∠DBA = ∠x ( these are the angles which have opposite sides)

In ΔDCB, BD = CD

∠DBC = ∠DCB = ∠y

In ΔABC we will use the angle sum property

∠ABC + ∠BCA + ∠CAB = 180°

2(∠x + ∠y) = 180°

∠x + ∠y = 90°

∠ABC = 90°

This means that ABC is the right angled triangle.