Given:ABCD is a rhombus.
To prove:DE congruent to BE.
In rombus, we know opposite angle are equal.
so, angle DCB = angle BAD
SINCE, ANGLE DCB= BAD
SO, In triangle DCA
angle DCA=angle DAC
similarly, In triangle ABC
angle BAC=angle BCA
since angle BCD=angle BAD
Therefore, angle DAC =angle CAB
so, opposite sides of equal angle are always equal.
so,sides DC=BC
Now, In triangle DEC and in triangle BEC
1. .DC=BC (from above)............(S)
2ANGLE CED=ANGLE CEB (DC=BC)....(A)
3.CE=CE (common sides)(S)
Therefore,DE is congruent to BE (from S.A.S axiom)
31.98, I am not 100% sure.
Answer:
<h3>
m∠BAC = 80° </h3>
Step-by-step explanation:
m∠BCD = 145° ⇒ m∠BCA = 180° - 145° = 35°
From ΔABC:
m∠ABC + m∠BCA + m∠BAC = 180°
65° + 35° + m∠BAC = 180°
m∠BAC = 180° - 100°
m∠BAC = 80°
Answer:
Step-by-step explanation:
(7+4) any expression 2
1km is 1000 m
1000m+125m-375m=750m
