Question

ΔABC and ΔDBC are two isosceles triangles on the same base BC (see figure). Show that ∠ABD =∠ACD.

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.

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

HOPE THIS ANSWER WILL HELP YOU..

Answer 2

Hello mate ☺

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$\bold\red{Solution:}$

AB=AC                  (Given)

It means that ∠ABC=∠ACB   (In triangle, angles opposite to equal sides are equal) .....(1)

BD=DC             (Given)

It means that ∠DBC=∠DCB    (In triangle, angles opposite to equal sides are equal)  ......(2)

Adding (1) and (2), we get

∠ABC+ ∠DBC=∠ACB+∠DCB

⇒∠ABD=∠ACD

I hope, this will help you.☺

Thank you______❤

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