Question

mathematics class 9 exercise 11.3 question 4 and 6
AD is an altitude of an isosceles triangle ABC in which AB=AC. show that
AD bosects BC
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Answer 1

Step-by-step explanation:

this is the meaningful and important explanation of the aforementioned question

Answer 2

Given :- AD is an altitude of an isosceles triangle ABC in which AB=AC. show that AD bisects BC ?

Solution :-

In ∆ABC we have,

→ AB = AC

so,

→ ∠ABC = ∠ACB { Angle opposite to equal sides are equal.}

or,

→ ∠ABD = ∠ACD --------- Eqn.(1)

also,

→ AD ⟂ BC { given. }

so,

→ ∠ADB = ∠ADC = 90° ----------- Eqn.(2)

now, in ∆ADB and ∆ADC we have,

→ ∠ADB = ∠ADC { from Eqn.(2) }

→ AD = AD { common. }

→ AB = AC { given. }

then,

→ ∆ADB ≅ ∆ADC { By RHS congruence rule. }

therefore,

→ DB = DC { By CPCT . }

hence, we can conclude that, AD bisect BC .

[ Note :- from Eqn.(1) and Eqn.(2) we can also use AA congruence rule .]

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