91Ó°ÊÓ

If the two sides of a triangle are congruent then the angles opposite to those sides are congruent.

Consider $\mathrm{Δ}ABD$, in which$ADâ‰\dots BD$ then by isosceles triangle theorem, $∠1â‰\dots ∠2$, that is

$m∠1=m∠2=35$.

The measure of exterior angle is equal to the sum of measure of two remote interior angles of a triangle.

From the given figure, it can be observed that $∠3$is exterior angle and $∠1\text{and}∠\text{2}$are remote exterior angles, such that,

$$\begin{gathered} m∠3=m∠1+m∠2 \\ =35+35 \\ =70 \end{gathered}$$

Consider $\mathrm{Δ}DBC$, in which$DBâ‰\dots DC$ then by isosceles triangle theorem, $∠4â‰\dots ∠5$, that is

$m∠4=m∠5$.

The sum of measures of all the angles of a triangle is 180.

Consider $\mathrm{Δ}DBC$, such that,

$$\begin{gathered} m∠5+m∠4+m∠3=180 \\ m∠5+m∠5+70=180 \\ 2m∠5=110 \\ m∠5=55 \end{gathered}$$

Substitute 35 for $m∠2$and 55 for $m∠5$.

$$\begin{gathered} m∠ABC=m∠2+m∠5 \\ =35+55 \\ =90 \end{gathered}$$

Therefore, the measure of$∠ABC$ is 90.

Consider $\mathrm{Δ}ABD$, in which$ADâ‰\dots BD$ then by isosceles triangle theorem, $∠1â‰\dots ∠2$, that is

$m∠1=m∠2=k$.

From the given figure, it can be observed that $∠3$is exterior angle and $∠1\text{and}∠\text{2}$are remote exterior angles then by exterior angle theorem,

$$\begin{gathered} m∠3=m∠1+m∠2 \\ =k+k \\ =2k \end{gathered}$$

Consider $\mathrm{Δ}DBC$, in which$DBâ‰\dots DC$ then by isosceles triangle theorem, $∠4â‰\dots ∠5$, that is

$m∠4=m∠5$.

Consider $\mathrm{Δ}DBC$then by angle sum theorem,

$$\begin{gathered} m∠5+m∠4+m∠3=180 \\ m∠5+m∠5+2k=180 \\ 2m∠5+2k=180 \\ m∠5+k=90 \\ m∠5=90−k \end{gathered}$$

Substitute $k$for $m∠2$and $90−k$for $m∠5$.

$$\begin{gathered} m∠ABC=m∠2+m∠5 \\ =k+90−k \\ =90 \end{gathered}$$

Therefore, the measure of$∠ABC$ is 90.