QUESTION IMAGE
Question
pretest: polygons\
statement\treason\
- \tdraw $\overline{bd} \perp \overline{ac}$.\tconstruction\
- \\t$\angle abc \cong \angle bdc$\tangles with the same measure are congruent.\
- \\t$\angle bca \cong \angle dcb$\treflexive property of congruence\
- \\t$\delta abc \sim \delta bdc$\taa criterion for similarity\
- \\t$\frac{bc}{dc} = \frac{ac}{bc}$\tcorresponding sides of similar triangles are proportional.\
- \\t$bc^2 = ac \times dc$\tcross multiplication\
- \\t$\angle abc \cong \angle adb$\tangles with the same measure are congruent.\
- \\t$\angle bac \cong \angle dab$\treflexive property of congruence\
- \\t$\delta abc \sim \delta adb$\taa criterion for similarity\
- \\t$\frac{ab}{ad} = \frac{ac}{ab}$\tcorresponding sides of similar triangles are proportional.\
- \\t\\t\
- \\t\\t\
- \\t\\t\
- \\t\\t\
- \\t\\t\
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$ab^2 + bc^2 = ac^2$\
reason: multiplication\
$ab^2 + bc^2 = ac \times ac$\
reason: segment addition\
$ab^2 = ac \times ad$\
reason: cross multiplication\
$ab^2 + bc^2 = ac(ad + dc)$\
reason: distributive property\
$ab^2 + bc^2 = ac \times ad + ac \times dc$\
reason: addition
Step1: Cross multiply Step10
Starting from $\frac{AB}{AD} = \frac{AC}{AB}$, cross multiply to get the next statement.
$AB^2 = AC \times AD$
Step2: Add Step6 and Step11
Add the equations from Step6 ($BC^2 = AC \times DC$) and Step11.
$AB^2 + BC^2 = AC \times AD + AC \times DC$
Step3: Factor right-hand side
Use distributive property on the right-hand side.
$AB^2 + BC^2 = AC(AD + DC)$
Step4: Replace $AD+DC$ with $AC$
Use segment addition: $AD + DC = AC$.
$AB^2 + BC^2 = AC \times AC$
Step5: Simplify right-hand side
Multiply $AC \times AC$ to get $AC^2$.
$AB^2 + BC^2 = AC^2$
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- $AB^2 = AC \times AD$
Reason: cross multiplication
- $AB^2 + BC^2 = AC \times AD + AC \times DC$
Reason: addition
- $AB^2 + BC^2 = AC(AD + DC)$
Reason: Distributive Property
- $AB^2 + BC^2 = AC \times AC$
Reason: segment addition
- $AB^2 + BC^2 = AC^2$
Reason: multiplication