Question

"Question 5 ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that ∠ABD = ∠ACD.

Class 9 - Math - Triangles Page 124"

Answer 1

Use the result that angles opposite to equal sides of a triangle are equal in both ∆ABC & ∆DBC and then add these results to show the required result.
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Solution:

Given:
ABC and DBC are two isosceles triangles.

To Prove:
∠ABD = ∠ACD

Proof:

In ΔABC,
AB =AC. (given)
∠ACB= ∠ABC......(1)
[Since angles opposite to equal sides of a triangle are equal]

In ∆ DBC,
DB=DC (given)
∠DCB = ∠DBC......(2)

[Since angles opposite to equal sides of a triangle are equal]

On adding eq 1 & 2 , We get
∠ACB+ ∠DCB= ∠ABC+ ∠DBC

∠ACD= ∠ABD
Hence, ∠ACD = ∠ABD

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Hope this will help you......

Answer 2

Hello!!✋

Here is your answer .

Given, that ABC and DBC are two isosceles triangles on the same base.

Proof that , < ABD = < ACD

proof :-
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°.° ABC isoscele triangle on the base.

.°. < ABC = < ACB ---------(1)

°.° DBC isoscele triangle on the base.

.°. < DBC = < DCB ----------(2)

From (1) and (2) we get

< ABC + < DBC = < ACB + < DCB

=> < ABD = < ACD

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