Find the value of x in given figure
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Answers
Answer:
1. Draw EF parallel to BC. Then angle
DFE = 80° because of equal
corresponding angles of parallel lines.
2. Drop a perpendicular line to BC from
A, hitting BC at G. Because of congruent
triangles ABC and AGC, angle BAG =
= angle CAG = 10°.
3. Now draw line FC, calling H the point
where line FC intersects line BE. Line AG
passes through point H, because of
symmetry.
4. Angle BAC = 60° since the other angles of the triangle BHC are both 60°.
5. BE = FC (because of corresponding
sides of congruent triangles ABC and
EBC). BH = HC (call BH b) because
triangle ABC is isosceles. So by subtraction, FH = HE.
6. Since angle OF = 60° (vertical angle
to BHC), and because FH = HE from 5,
angle FHE = angle OF = 60°, so triangle
HE is equilateral. Thus, FE = FH = HE.
Call each of those sides a.
7. Now AF = AE (because AB = AC and
FB = EC, by subtraction AF = AE).
8. Because triangle ABC is isosceles, AE = BE = b + a. Thus, AF = BE = b + a (since
= AF = AE from 7).
9. BE = FC (congruent triangles BEC and
BFC), so AF = FC, since AF = BE from 8.
10. Triangle ABC is congruent to triangle CFD because AF = FC; angle OF = 140°
= angle CFD; angle DCF = 10° = angle FAH. Thus, corresponding sides of the
congruent triangles AFH and triangle CFD are equal, so FH = FD. But FH = FE
from 6, so FE = FD.
11. Science FOR = FD, angle ADE = angle FED
and since angle DFE = 80° from (1), angle
FDE = 50° = angle FED.
12.
So angle DBC = 30° (because DBC = 80° and BCD = 70°), so by subtracting angle BDE = 50° from angle BDC, angle CDE = 20°
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Answer:
angle DEA will be 30 degree