Question

1. isosceles $\triangle abc$, base $\overline{ac}$

$\overline{bd} \perp$ bisector of $\overline{ac}$
$ad = x + 1$; $dc = 2x - 4$
$bd = x + 3$
find $ab$.

Explanation:

Step1: Since BD is the bisector of AC in isosceles triangle ABC, AD = DC.

So we set up the equation: $x + 1 = 2x - 4$

Step2: Solve the equation for x.

Subtract x from both sides: $1 = x - 4$
Add 4 to both sides: $x = 5$

Step3: Find the lengths of AD and BD.

AD = $x + 1 = 5 + 1 = 6$
BD = $x + 3 = 5 + 3 = 8$

Step4: Use the Pythagorean theorem in right triangle ABD to find AB.

In right triangle ABD, $AB^2 = AD^2 + BD^2$
So $AB^2 = 6^2 + 8^2 = 36 + 64 = 100$
Take the square root of both sides: $AB = \sqrt{100} = 10$

Answer:

10