Prove: If triangle ABD is congruent to triangle DEA, then angle CBD is congruent to angle FEA
and
Given: Triangle ABD is congruent to Triangle DEA; Triangle ACD is congruent to Triangle DFA
Prove: Triangle AFE is congruent to Triangle DCB.
Any help is appreciated.....
2007-02-06 14:36:44 · 4 answers · asked by hot4manny 2 in Science & Mathematics ➔ Mathematics
The diagram applies to both of the proofs...
http://img122.imageshack.us/img122/3179/prooftr5.jpg
2007-02-06 14:55:42 · update #1
and if possible in the two-column form....
2007-02-06 15:26:18 · update #2
1st one:
(Technically, from the drawing and stated question, there is nothing to require the sides that pair up be the obvious ones, but, I will take them to be or this becomes either unsolvable or pretty involved with laying out all the possible cases. More so than I have time for! I also assume that ABC and DEF are line segments - so "angle" ABC is 180.)
Since ABD and DEA are congruent, I will take angles B and E as being congruent and for side AD to be paired with itself in the triangle congruency. So, AE and BD are congruent and AB and ED are congruent. If one draws the segment BE and considers triangles AEB and BDE, we find they are congruent as well. That makes AD and BE congruent as all four triangles are congruent. So, ABDE has to be a rectangle. If so, CBD and FEA are supplements to right angles, therefore right angles themselves. And therefore congruent.
2nd one:
Use the congruency of the first pair of triangles to pair up sides and angles. Then use the congruency of the second pair of triangles to pair up sides and angles. Pairing the angles shows angle ADC and angle FAD to be congruent and angles ADB and EAD to be congruent. Subtracting the smaller angles from the larger ones, leaves the angles FAE and BDC and they must be congruent since they are the leftovers from similar subtractions. Now you have side-angle-side (BD-AE, BDC-FAE, DC-FA) congruency.
You have to use the subtraction approach, either with the angles as I did, or with the segments BC and EF.
2007-02-06 14:42:37 · answer #1 · answered by roynburton 5 · 0⤊ 0⤋
We need to know where points F and C are. Otherwise we could put them anywhere and keep changing the values of the angles.
2007-02-06 14:47:30 · answer #2 · answered by Anonymous · 0⤊ 0⤋
well if the angles are congruent and they hav different labels dat cant work!!
like where did the vertex f *** from its impossible!!
2007-02-06 14:41:41 · answer #3 · answered by nahir_lozza 1 · 0⤊ 0⤋
The first help is to get your problem statements right.
In the first problem, you havent defined points C and F, so there is nothing we can do for you.
2007-02-06 14:47:47 · answer #4 · answered by cattbarf 7 · 0⤊ 0⤋