Question

prove that the bisector of the verticle angle of an isosceles triangle bisects the base at right angles????

Answer 1

Step-by-step explanation:

Let ABC is the isosceles triangle where AB = AC and A is the vertical angle (refer to attached figure). AD is the bisector of ∠A which intersects BC at point D.

Given : AB = AC, ∠BAD = ∠CAD

To Prove: BD = CD and ∠ADB = ∠ADC = 90°

Proof:

                        In ΔADB and ΔADC

                             AB = AC                      [Given]

                        ∠BAD = ∠CAD                [Given]

                             AD = AD                     [ Common ]

                    ∴ ΔADB ≅ ΔADC               [by SAS]

                         ∴ DB = DC                     [ CPCT] --- ( i )

                       ∠ADB = ∠ADC                [ CPCT ] --- ( ii )

but ∠ADB +∠ADC = 180° [ Linear Pair]

                      ∠ADB + ∠ADB = 180°   [∵∠ADB = ∠ADC]

                                  2∠ADB = 180°

                                    ∠ADB = 90° ---- ( iii )

From equations ( i ) ( ii ) and ( iii ), we can say that AD is bisecting base BC at right angles.