Question

"Question 2 AD is an altitude of an isosceles triangles ABC in which AB = AC. Show that (i) AD bisects BC (ii) AD bisects ∠A.

Class 9 - Math - Triangles Page 128"

Answer 1

Congruence of triangles:
Two ∆’s are congruent if sides and angles of a triangle are equal to the corresponding sides and angles of the other ∆.

In Congruent Triangles corresponding parts are always equal and we write it in short CPCT i e, corresponding parts of Congruent Triangles.

It is necessary to write a correspondence of vertices correctly for writing the congruence of triangles in symbolic form.

Criteria for congruence of triangles:


RHS(right angle hypotenuse side):
In two right angled triangles, the hypotenuse and one side of One triangle are equal to the hypotenuse and one side of the Other triangle, then the two Triangles are congruent.

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Use RHS congruence rule to show ΔABD ≅ ΔACD and then CPCT to prove given parts

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Given:
∆ABC is an isosceles triangle &
AD is an altitude & AB = AC.


To Show:

(i) AD bisects BC
(ii) AD bisects ∠A.


Proof:

(i) In ΔABD and ΔACD,
∠ADB = ∠ADC = 90°
AB = AC (Given)
AD = AD (Common)

Therefore,
ΔABD ≅ ΔACD
(by RHS congruence rule)

Then,
BD = CD (by CPCT)

Hence, AD bisects BC.


ii) As AD bisects BC , then ∠BAD = ∠CAD (by CPCT)

Thus, AD bisects ∠A.


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Hope this will help you.....

Answer 2

Hello !

I can help you , here is your answer

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Solution:-
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Given us, AD is a altitude of an isosceles triangles ABC in which AB = AC.

we need to prof that ,

( I ) AD bisects BC

( ii ) AD bisects A

now ,

Prof :

From ∆ ADB and ∆ ADC we get,

( both are right angles )

Hypotenuse AB = Hypotenuse AC ( Given )

Figure AD = Figure AD ( General figure )

.°. ∆ ADB =~ ∆ ADC ( by congruence rule of RHS )

.°, BD = CD ( by CPCT )

=> AD , bisecor of BC


( ii ) ∆ ADB =~ ∆ ADC ( Already probed )

.°. < BAD = < CAD ( By CPCT )

=> AD besects < A

Hopes I helped

thanks


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