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Standard IX

Mathematics

If Two Angles of a Triangle Are Congruent Then the Sides Opposite to Them Are Congruent

In an isoscel...

Question

In an isosceles ∆ABC, AB = AC and the bisectors of are joint.
Prove that: (i) OB = OC (ii) ∠OAB = ∠OAC

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Solution

In $△ A B O a n d △ A C O$ , we have:
$AB = AC (Given) AO = AO (Common) ∠ ABO = ∠ ACO (In an isoscles triangle, opposite angles are equal) △ ABO ≅ △ ACO (SAS criterion)$
⇒ OB = OC (CPCT)
⇒ $∠ O A B = ∠ O A C (CPCT)$
Hence proved.

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Question 1 (i)
In an isosceles triangle ABC, with AB = AC, the bisectors of $∠ B a n d ∠ C i n t e r s e c t$
each other at O. Join A to O. Show that:
OB = OC

O is any point in the interior of $Δ A B C .$

Prove that

$(i) A B + A C > O B + O C$

$(i i) A B + B C + C A > O A + O B + O C$

$(i i i) O A + O B + O C > \frac{1}{2} (A B + B C + C A)$

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In an isosceles ∆ABC, AB = AC and the bisectors of are joint. Prove that: (i) OB = OC (ii) ∠OAB = ∠OAC

In △ABO and △ACO , we have: AB=AC (Given)AO=AO (Common)∠ABO=∠ACO (In an isoscles triangle, opposite angles are equal)△ABO≅△ACO (SAS criterion) ⇒ OB = OC (CPCT) ⇒ ∠OAB=∠OAC (CPCT) Hence proved.

In an isosceles ∆ABC, AB = AC and the bisectors of are joint.
Prove that: (i) OB = OC (ii) ∠OAB = ∠OAC

In $△ A B O a n d △ A C O$ , we have:
$AB = AC (Given) AO = AO (Common) ∠ ABO = ∠ ACO (In an isoscles triangle, opposite angles are equal) △ ABO ≅ △ ACO (SAS criterion)$
⇒ OB = OC (CPCT)
⇒ $∠ O A B = ∠ O A C (CPCT)$
Hence proved.