Can you help me in my math in givens and proves?

Question

If AC and BD intersects at E,
this is the picture of the question.
http://aycu39.webshots.com/image/24278/2000688765522279865_rs.jpg

Answer 1

given Angle 1 = Angle 2
given angle D = angle C
Because M is the midpoint of DC, DM=DC.
Consider triangles BCM and ADM
Because we have two similar angles and one similar side, the triangles are similar.
Therefore BM=AM
Consider triangles BDM and ADM.
Because we have two similar sides and one similar angle, the triangles are similar.
Therefore DB=CA.

Answer 2

to prove DB =~ CE, we need to prove Triange ACM =~ Triangle BDM

Let's call F is the point where DE intersect AM
let's call point G is the pont where CE intersect BM

1) AC and BD intersects at E,
2) Goal: prove

we know that

3) DM = MC by Midpoint Postulate

4) Triangle FMD = Triangle GMC by ASA

5) FM = GM by CPCTC

6) Goal: prove
<1 + <2 + <1 +
we know that <1 = <2
<1 +
7) since

8) < BGE = <2 by Vertical Angle Theorem

9)
10)
11) Trianle AEF = Triangle BEG by SAS

12)
13) Triange BDM = Triangle ACM by AAS

14) BD = AC by CPCTC

Answer 3

I'm sorry but I don't know which angles 1 and 2 refer to, so I'm just going to assume, and I'm not sure how to write this in a formal way (meaning in steps), so I will write it and you can figure that out! Since

Answer 4

alright
i just have a question
which is angle 1 and which is angle 2
they r just in the middle, they rn't in any particular angle

but to prove DB~= to AC
u need to first prove that triangle CAM is ~= to triangle DBM
u know DM~=MC
and that the angles r the same
u should know the postulates of the triangles

hope i helped

Answer 5

to your photograph in quantity a million, BH isn't existant. although, i think which you advise that the line containing H and EG are parallel. in any case, use the thought approximately parallel strains shrink via a transversal. In the style of case, the corresponding angles are congruent. And if I interpreted your question properly, the <3 ~= <2 isn't appropriate. For quantity 2, ST perpendicular PR potential attitude RTS is a maximum suitable attitude. for the reason that triangle TRS and QRS the two comprise maximum suitable angles, they are maximum suitable triangles. and there's a theorem declaring that maximum suitable triangles are congruent in the event that they have 2 congruent factors. we are given RT congruent to RQ. additionally, RS, it extremely is a edge of the two triangles, is congruent to itself. for this reason, the triangles aforementioned are congruent.

Answer 6

no! there is a worm!