In the figure below, ∠dec ≅ ∠dce, ∠b ≅ ∠f, and df congruent to bd. point c is the point of intersection between

4 answers
Question:

in the figure below, ∠dec ≅ ∠dce, ∠b ≅ ∠f, and df congruent to bd. point c is the point of intersection between ab and bd while point e is the point of intersection between ag and df. prove δabc ≅ δgfe.
what i have right now is;
∠b ≅ ∠f given
cb ≅ ef df ≅ bd, (given), df - ed ≅ bd - cd
i'm trying to prove ∠dce ≅ ∠acb and ∠dec ≅ ∠gef so that i can use asa to prove δabc ≅ δgfe, but i don't know how to do that and this is very urgent so

[tex]in the figure below, ∠dec ≅ ∠dce, ∠b ≅ ∠f, and df congruent to bd. point c is the point of in[/tex]


Answers

answer:

its a

step-by-step explanation:

answer:

(ex: a = b and b = c, then a = c) ab//cd and p is any point. prove that angle abp+angle bpd + angle cdp=360^0. (the fig. in the figure below, ∠dec ≅ ∠dce, ∠b ≅ ∠f, and segment df is congruent to segment bd. point c is the point of intersection between segment ag and segment bd, while point e is the point of .

step-by-step explanation:

Step-by-step explanation:

Angle B and angle F are known to be equal.

Since angles DCE and DEC are equal, angles ACB and GEF are also equal (vertically opposite angles)

When two angles are equal, the third angle has to be equal too because total angle of both triangles is the same, i.e 180

Now when angles are equal, if we just show that lengths BC ans EF are also equal, it would be sufficient to say that the two triangles are congruent.

Since triangle DCE is an isosceles triangle (because two angles same),

Length CD = ED

We know that

Length of BD = Length FD

Length CD = Length ED

Length of BC = BD - CD

Length of EF = FD - ED

Therefore,

Length BC = Length EF

When all angles are equal, and even one pair of corresponding lengths is equal.. the triangles are congruent

∠ABC ≅ ∠DEC

∠DEC = ∠GEF

thus:

∠ABC = ∠GEF

∠GFE ≅ ∠DCE

∠DCE = ∠ACB

thus:

∠ACB = ∠GFE

If two of their angles are equal, then the third angle must also be equal, because angles of a triangle always add to make 180°.

thus: ΔABC ∼ ΔGEF by AA similar triangle postulate

I hope this helps

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