Question

.В равнобедренном треугольнике АВС с основанием АС проведена биссектриса ВД. Докажите, что ∆АВД=∆СВД.

Answer 1

Биссектриса ВД, делит на равные углы угол В, т.е. угол АВД = углу СБД
Биссектриса ВД проходит к основанию АС, делит  на пополам её АД = ДС
ВД - общая сторона, АВ = ВС, АД =ДС следовательно, треугольники АВД= тр. СВД

Answer 2

Proof that ∆АВД = ∆СВД in an isosceles triangle АВС with base АС and bisector ВД

To prove that ∆АВД is congruent to ∆СВД in an isosceles triangle АВС with base АС and bisector ВД, we can use the following steps:

1. Given: Isosceles triangle АВС with base АС and bisector ВД. 2. To prove: ∆АВД = ∆СВД.

Proof:

Step 1: In triangle АВС, we have: - Side AB = Side AC (Given that triangle АВС is isosceles). - Angle ВАС = Angle САВ (By definition of an isosceles triangle).

Step 2: In triangle ВАD, we have: - Angle ВАD = Angle ВАС/2 (By definition of a bisector).

Step 3: In triangle СВD, we have: - Angle СВD = Angle САВ/2 (By definition of a bisector).

Step 4: From Step 1, we know that: - Angle ВАС = Angle САВ.

Step 5: Combining Steps 2, 3, and 4, we can conclude that: - Angle ВАD = Angle СВD.

Step 6: From Step 1, we know that: - Side AB = Side AC.

Step 7: By the Side-Angle-Side (SAS) congruence criterion, we can conclude that: - ∆АВД = ∆СВД.

Therefore, we have proved that ∆АВД is congruent to ∆СВД in an isosceles triangle АВС with base АС and bisector ВД.