Plz prove this triangle congruence.

Mathematics

1 answer:

alex41 [277]4 years ago

8 0

Answer:

ΔDBE≅ΔQAP (by RHS criteria)

Step-by-step explanation:

Given that, $PQ=DE$ , $PB=AE$ , $QA$ ⊥ $PE$

and $DB$ ⊥ $PE$

⇒∠PAQ=90° and ∠EBD=90°(definition of perpendicular lines)

Its given that PB=AE,

subtracting AB on both sides,

we get: PB-AE=AB-AE

⇒PA=EB (equals subtracted from equals, the remainders are equals)

Therefore, ΔDBE≅ΔQAP (by RHS criteria)

conditions for congruence:

∠DBE=∠QAP=90°(right angle)
PQ=ED(hypotenuse)
PA=EB(side)

So, ∡D=∡Q(as congruent parts of congruent triangles are equal)

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3 years ago

MATH HELP!

Irina18 [472]

You have the correct answer. Nice work. If you need to see the steps, then see below

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First we need to find the midpoint of H and I
The x coordinates of the two points are -4 and 2. They add to -4+2 = -2 and then cut that in half to get -1

Do the same for the y coordinates: 2+4 = 6 which cuts in half to get 3

So the midpoint of H and I is (-1,3). The perpendicular bisector will go through this midpoint

---------------------

Now we must find the slope of segment HI

H = (-4,2) = (x1,y1)
I = (2,4) = (x2,y2)
m = (y2 - y1)/(x2 - x1)
m = (4 - 2)/(2 - (-4))
m = (4 - 2)/(2 + 4)
m = 2/6
m = 1/3
Flip the fraction to get 1/3 ---> 3/1 = 3
Then flip the sign: +3 ----> -3

So the slope of the perpendicular bisector is -3

-----------------------

Use m = -3 which is the slope we found
and (x,y) = (-1,3), which is the midpoint found earlier
to get the following
y = mx+b
3 = -3*(-1)+b
3 = 3+b
3-3 = 3+b-3
0 = b
b = 0

So if m = -3 and b = 0, then y = mx+b turns into y = -3x+0 and it simplifies to y = -3x

So that confirms you have the right answer. I've also used GeoGebra to help confirm the answer (see attached)