In the following figure, Show that △BAD≅△BCD. *

Mathematics

1 answer:

Otrada [13]3 years ago

3 0

Answer and Step-by-step explanation:

Side AD is congruent to side AC by the angle bisector theorem.

Angle A is congruent to angle C.

Side AB is congruent to side BC by the angle bisector theorem.

Triangle BAD is congruent to triangle BCD by SAS congruence theorem.

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Please help !!!!!!!!!

Anettt [7]

the answer will be -19n+25. All you do is put 19 into the g spot and distribute from that point on into the parentheses. so -19 to n and multiply -19 and -1. after that you get -19 +19+6. you will then add like termsso 19 and 6 is 25. Final answer is -19n+25. Hope its right. Good Luck !

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4 years ago

Prove the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus. use the distance formula

riadik2000 [5.3K]

In order to prove this, we have to put the trapezoid to the coordinate system. In the attached photo you can see how it has to be put. The coordinates for the vertices of trapezoid written according to the midpoint principle. By using the distance between two points formula, we can find the coordinates for the vertices of the rhombus.
$D_{x} = \frac{-2b-2a}{2}=-b-a$ and $D_{y}= \frac{2c+0}{2}=c$ . The coordinates of D is $(-b-a, c)$
$E_{x} = \frac{-2b+2b}{2}=0$ and $E_{y}= \frac{2c+2c}{2}=2c$ . The coordinates of E is $(0,2c)$
Since we have the reflection in this graph, the coordinates of F is $(b+a, c)$
And the coordinates of G is (0,0).

Using the distance formula, we can find that
$DE= \sqrt{(-b-a)^{2}+ c^{2}}=\sqrt{(b+a)^{2} + c^{2}}$
$EF= \sqrt{(b+a)^{2} + c^{2} }$
$GF= \sqrt{(b+a )^{2} + c^{2} }$
$DG= \sqrt{(b+a)^{2}+ c^{2}$

Since all the sides are equal this completes our proof. Additionally, we can find the distances of EG and DF in order to show that the diagonals of this rhombus are not equal. So that it is not a square, but rhombus.