M∠ABC when m∠BAC = 70° and ΔABC is an isosceles triangle with

m∠QPR when m∠QRP = 30°
and ΔPQR is an isosceles triangle with

m∠BDE when m∠BAC = 45°
and points D and E are the midpoints of
and , respectively, in ΔABC
55°
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Given:

There are 4 statements and we need to find the answer of each statement.

Statement 1:

m ∠ABC + m ∠BAC + m ∠ACB when Δ ABC is an isosceles triangle

Sum of all interior angles of a triangle is always equal to 180°.

⇒  m ∠ABC + m ∠BAC + m ∠ACB = 180°

Therefore, the correct option is (B).

Statement 2:

m∠ABC when m∠BAC = 70° and ΔABC is an isosceles triangle

From the triangle ABC, ∠A = 70, ∠B = ∠C(Isosceles triangle property)

Let ∠B = ∠C = 'x'.

Sum of all interior angles of a triangle is always equal to 180°.

⇒  m ∠ABC + m ∠BAC + m ∠ACB = 180°

m ∠ABC = ∠B = 55° [Option (A)]

Statement 3:

m∠QPR when m∠QRP = 30° and ΔPQR is an isosceles triangle.

From the triangle QPR, ∠R = 30°, QP ≅ QR

Now, from isosceles triangle definition, the angles opposite to the congruent sides are also congruent. Therefore,

∠P = ∠R = 30°

Now, m∠QPR is same as ∠P. So,  m∠QPR = 30° [Option D].

Statement 4:

m∠BDE when m∠BAC = 45° and points D and E are the midpoints of AB and BC, respectively, in ΔABC

From the figure shown below,

DE is the midsegment of sides AB and BC. Therefore, from midsegment theorem, DE || AC (third side).

Therefore, for two parallel line DE and AC cut by transversal AB,

m∠BDE = m∠BAC (Corresponding angles are congruent)

m∠BDE = 45°

Explanation:

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explanation:

answer; i believe the correct answer is revolution)