The length of DE = 4.5 units and m∠CAB = 45°
Explanation:
Given:
AB = BC
D and E are midpoints of AB and BC.
Hence, DE will be the midsegment of CA
We will apply the formula for midsegment of triangle to get DE
Midsegment formula:
[tex]\begin{gathered} DE\text{ = }\frac{1}{2}(CA) \\ \end{gathered}[/tex][tex]\begin{gathered} CA\text{ = 9 units} \\ DE\text{ = }\frac{1}{2}(9) \\ DE\text{ = 4.5 units} \end{gathered}[/tex]
m∠ABC = 90°
Since AB = BC, the triangle is an isosceles triangle.
As a result, the base angles will also be the same
∠CAB = ∠BCA
∠ABC + ∠CAB + ∠BCA = 180° (sum of angles in a triangle)
∠ABC + ∠CAB + ∠CAB = 180°
90 + 2∠CAB = 180
2∠CAB = 180 - 90
2∠CAB = 90
Divide both sides by 2:
∠CAB = 90/2
∠CAB = 45°
If the length CA is 9 units, the length of DE is 4.5 units and m∠CAB is 45°
