Question

In triangle ABC,AB=ACand line AO divides the base BC into equal parts.Prove that triangle AOB is congruent to triangle AOC​

Answer 1

Answer:

Ans. In quadrilateral ABCD we have

Ans. In quadrilateral ABCD we have AC = AD

Ans. In quadrilateral ABCD we have AC = AD and AB being the bisector of ∠A.

Ans. In quadrilateral ABCD we have AC = AD and AB being the bisector of ∠A. Now, in ΔABC and ΔABD,

Ans. In quadrilateral ABCD we have AC = AD and AB being the bisector of ∠A. Now, in ΔABC and ΔABD, AC = AD

Ans. In quadrilateral ABCD we have AC = AD and AB being the bisector of ∠A. Now, in ΔABC and ΔABD, AC = AD[Given]

Ans. In quadrilateral ABCD we have AC = AD and AB being the bisector of ∠A. Now, in ΔABC and ΔABD, AC = AD[Given] AB = AB

Ans. In quadrilateral ABCD we have AC = AD and AB being the bisector of ∠A. Now, in ΔABC and ΔABD, AC = AD[Given] AB = AB[Common]

Ans. In quadrilateral ABCD we have AC = AD and AB being the bisector of ∠A. Now, in ΔABC and ΔABD, AC = AD[Given] AB = AB[Common]∠CAB = ∠DAB [∴ AB bisects ∠CAD]

Ans. In quadrilateral ABCD we have AC = AD and AB being the bisector of ∠A. Now, in ΔABC and ΔABD, AC = AD[Given] AB = AB[Common]∠CAB = ∠DAB [∴ AB bisects ∠CAD] ∴ Using SAS criteria, we have

Ans. In quadrilateral ABCD we have AC = AD and AB being the bisector of ∠A. Now, in ΔABC and ΔABD, AC = AD[Given] AB = AB[Common]∠CAB = ∠DAB [∴ AB bisects ∠CAD] ∴ Using SAS criteria, we have ΔABC ≌ ΔABD.

Ans. In quadrilateral ABCD we have AC = AD and AB being the bisector of ∠A. Now, in ΔABC and ΔABD, AC = AD[Given] AB = AB[Common]∠CAB = ∠DAB [∴ AB bisects ∠CAD] ∴ Using SAS criteria, we have ΔABC ≌ ΔABD. ∴ Corresponding parts of congruent triangles (c.p.c.t) are equal.

Ans. In quadrilateral ABCD we have AC = AD and AB being the bisector of ∠A. Now, in ΔABC and ΔABD, AC = AD[Given] AB = AB[Common]∠CAB = ∠DAB [∴ AB bisects ∠CAD] ∴ Using SAS criteria, we have ΔABC ≌ ΔABD. ∴ Corresponding parts of congruent triangles (c.p.c.t) are equal. ∴ BC = BD.