Question

AD is an altitude of an isosceles triangle ABC in which A = AC. Show that
(i)AD bisects BC
(ii)AD bisect angle A

Answer 1

In ∆ABD and ∆ACD
angle ADB=angle ADC (each 90 degree)
AB=AC (given)
AD=AD( common)
therefore, ∆ABD =~ ∆ACD by RHS rule

therefore, BD=CD
THEREFORE, AD bisects BC

Answer 2

Hello mate ^_^

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Solution (i)

Given: ∆ABC is isosceles with AB=AC and AD is perpendicular to BC.

We need to prove that BD=CD.

In ∆ABD and ∆ACD, we have

AB=AC            (Given)

∠ADB=∠ADC          (Each given equal to 90°)

AD=AD              (Common)

Therefore, by RHS congruence rule, ∆ABD≅∆ACD

Hence, we have BD=CD           (Corresponding parts of congruent triangles are equal).

Solution (ii)

We have proved above that ∆ABD≅∆ACD.

It means that ∠BAD=∠CAD          (Corresponding parts of congruent triangles are equal).

∠BAD=∠CAD means that AD bisects ∠A.

hope, this will help you.

Thank you______❤

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