Question

In a AABC, AB = AC. If D is the mid-point of the side BC, prove that: (i) AD is perpendicular toBC (ii) AD bisects angle BAC.​

Answer 1

Step-by-step explanation:

In △ABD and △ACD,

∠BAD=∠CAD (Given)

AD=AD (Common)

AB=AC (Given)

Thus, △ABD≅△ACD (SAS rule)

Hence, BD=CD (By cpct)

∠ADB=∠ADC=x (By cpct)

∠ADB+∠ADC=180

x+x=180

x=90

∘

Thus, ∠ADB=∠ADC=90

∘

Hence, AD is perpendicular bisector of BC

Answer 2

Given:- AB=AC , D is the midpoint of the side BC.

To prove :- AD is perpendicular to BC

Proof:-

In △ABD and △ACD,

∠BAD=∠CAD (Given)

AD=AD (Common)

AB=AC (Given)

Thus, △ABD≅△ACD (SAS rule)

Hence, BD=CD (By cpct)

∠ADB=∠ADC=x (By cpct)

∠ADB+∠ADC=180

x+x=180

x=90°

Thus, ∠ADB=∠ADC=90°

Hence, AD is perpendicular bisector of BC

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