91Ó°ÊÓ

The labelled diagram is:

From the diagram it can be noticed that:

$∠BAD=∠BAC+∠CAD$

Therefore, it can be obtained that:

$$\begin{gathered} ∠BAD=∠BAC+∠CAD \\ =50°+40° \\ =90° \end{gathered}$$

Therefore, the measure of the angle $∠BAD$is $90°$.

The sum of the angles of a triangle is $180°$.

Therefore, the sum of the angles of a triangle $\mathrm{Δ}ABD$is $180°$.

Therefore, it can be obtained that:

$\begin{array}{c}∠BAD+∠ABD+∠ADB=180°\\ 90°+55°+∠2=180°\\ 145°+∠2=180°\end{array}$

$$\begin{gathered} ∠2=180°−145° \\ ∠2=35° \end{gathered}$$

Therefore, the measure of the angle $∠2$is $35°$.

The sum of the angles of a triangle is $180°$.

Therefore, the sum of the angles of a triangle $\mathrm{Δ}ACD$is $180°$.

Therefore, it can be obtained that:

$$\begin{gathered} ∠CAD+∠ACD+∠ADC=180° \\ 40°+∠2+∠1=180° \\ 40°+35°+∠1=180° \\ 75°+∠1=180° \\ ∠1=180°−75° \\ ∠1=105° \end{gathered}$$

Therefore, the measure of the angle $∠1$is $105°$.

In the diagram, $m∠1=105$and $m∠2=35$.