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3 years ago

10

Given AD CB ADB CBD Prove: ∆ADB ≅ ∆CBD

2 answers:

algol [13]3 years ago

8 0

Answer:

<h2>SAS postulate.</h2>

Step-by-step explanation:

Notice that sides AD and BC are congruent, also DB is a common side to both triangles. At this point, we already have two corresponding congruent sides.

Additonally, angles ADB and CBD are congruent. In summary, we have two pair of corresponding sides congruent, and one pair of corresponding angles congruent (which is formed by the congruent sides).

Based on all the given information, by SAS postulate triangles ADB and CBD are congruent.

Andrei [34K]3 years ago

7 0

We will use the SAS (SIDE-ANGLE-SIDE) to prove that two triangles are congruent.

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3 years ago

Find the coordinates of the point 7/10 of the way from A to B. a=(-3,-6) b=(12,4)

Artemon [7]

Answer:

The coordinates of $M$ are $x = \frac{15}{2}$ and $y = 1$ .

Step-by-step explanation:

Let be $A = (-3,-6)$ and $B = (12, 4)$ endpoints of segment AB and $M$ a point located 7/10 the way from A to B. Vectorially, we get this formula:

$\overrightarrow {AM} = \frac{7}{10}\cdot \overrightarrow {AB}$

$\vec M - \vec A = \frac{7}{10}\cdot (\vec B - \vec A)$

By Linear Algebra we get the location of $M$ :

$\vec M = \vec A + \frac{7}{10}\cdot (\vec B - \vec A)$

$\vec M = \vec A +\frac{7}{10}\cdot \vec B - \frac{7}{10}\cdot \vec A$

$\vec M = \frac{3}{10}\cdot \vec A + \frac{7}{10}\cdot \vec B$

If we know that $\vec A = (-3,-6)$ and $\vec B = (12, 4)$ , then:

$\vec M = \frac{3}{10}\cdot (-3,-6)+\frac{7}{10}\cdot (12,4)$

$\vec M = \left(-\frac{9}{10},-\frac{9}{5} \right)+\left(\frac{42}{5} ,\frac{14}{5} \right)$

$\vec M =\left(-\frac{9}{10}+\frac{42}{5} ,-\frac{9}{5}+\frac{14}{5} \right)$

$\vec M = \left(\frac{15}{2} ,1\right)$

The coordinates of $M$ are $x = \frac{15}{2}$ and $y = 1$ .

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3 years ago

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3 years ago

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