Question

1. if cd = 4 and the perimeter of △abc is 23, what is the perimeter of △abe?

Explanation:

Step1: Prove $\triangle ADE\cong\triangle ADC$

Since $AD\perp EC$, $\angle ADE=\angle ADC = 90^{\circ}$. Also, $AD = AD$ (common side) and $AE=AC$ (given). By the Hypotenuse - Leg (HL) congruence criterion for right - angled triangles, $\triangle ADE\cong\triangle ADC$. So, $DE = DC$.

Step2: Express the perimeter of $\triangle ABC$ and $\triangle ABE$

The perimeter of $\triangle ABC$ is $P_{ABC}=AB + BC+AC=AB+(BD + DC)+AC$.
The perimeter of $\triangle ABE$ is $P_{ABE}=AB + BE+AE=AB+(BD + DE)+AE$.

Step3: Substitute the equal - length segments

Since $DE = DC$ and $AE = AC$, we substitute these into the perimeter formula of $\triangle ABE$.
$P_{ABE}=AB+(BD + DC)+AC$.
Given that $P_{ABC}=AB+(BD + DC)+AC = 23$ and $CD = 4$ (but the value of $CD$ is not needed for the final calculation as the perimeters of $\triangle ABC$ and $\triangle ABE$ have the same expression in terms of side lengths).

Answer:

The perimeter of $\triangle ABE$ is $31$.