Given: ab ≅ bc bd − median of δabc, m∠abd = 40° find: m∠abc, m∠bdc

m∠ABC=80° and m∠BDC=90°

Step-by-step explanation:

Given the ΔABC in which AB ≅ BC, m∠ABD = 40° and BD is median of ΔABC.

we have to find the measure of angle ∠ABC and ∠BDC.

As the median of isosceles triangle split the angle at the vertex into two equal parts i.e ∠ABC is twice the angle ∠ABD

⇒

Also the median of isosceles triangle is perpendicular to the opposite side i.e to the base. Here, BD is perpendicular to AC

⇒  ∠BDC=90°

Therefore, the measure of angle ABC and BDC is 80° and 90° respectively.

4x and 6y

im not sure though