Question

ABC and DBC are two isosceles triangles in the same base BC show that <ABC =<ACD​

Answer 1

Given :

• ABC and DBC are two isosceles triangle on the same base BC.

To Prove :

• $\displaystyle{\sf{ \angle ABD = \angle ACD}}$

Solution :

$\displaystyle{\sf{ \because ABC\:is\:an\:isosceles\:triangle\:on\:the\:base\:BC.}}$

$\displaystyle{\sf{ \therefore \angle ABC = \angle ACB\:\:\:\:\:....(1)}}$

$\displaystyle{\sf{ \because DBC\:is\:an\:isosceles\:triangle\:on\:the\:base\:BC.}}$

$\displaystyle{\sf{ \therefore \angle DBC = \angle DCB\:\:\:\:\:....(2)}}$

Adding the corresponding sides of (1) and (2), we get

$\displaystyle{\sf{ \angle ABC + \angle DBC = \angle ACB + \angle DCB}}$

$\large{\boxed{\sf{\red{ \angle ABD = \angle ACD}}}}$

Hence, proved.

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Answer 2

Given :

• ABC and DBC are two iscoceles triangles.

To Prove :

• ∠ABD = ∠ACD

Solution :

In Δ ACD and ΔABD :

$: \implies\tt{AC=AB(Given)} \\ </p><p> \\ </p><p> : \implies\tt{DC=BD(Given)} \\ \\ </p><p></p><p> : \implies\tt{AB=AB(Common)}$

So , By SSS rule Δ ACD ≅ Δ ABD...

Now :

$: \implies\tt{\angle{ABD}=\angle{ACD} \: \: (By\:CPCT\:Rule)} </p><p>$

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SAS Rule :

• If three sides of one triangle are equal to the three sides of another triangle then the two triangles are congruent.

Congruence of Triangle :

• Two are congruent if the sides and angles of one triangle are equal to corresponding sides and angles of the other triangle according to some other rules.

CPCT means Corresponding Parts of Congruent Triangle.

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