In ΔABC (m∠C = 90°), the points D and E are the points where the angle bisectors of ∠A and ∠B intersect respectively sides BC an d AC . Point G ∈ AB so that DG ⊥ AB and H ∈ AB so that EH ⊥ AB .Prove m∠HCG = 45° (with statement reason preferably).
1 answer:
This is a little long, but it gets you there.
ΔEBH ≅ ΔEBC . . . . HA theorem EH ≅ EC . . . . . . . . . CPCTC ∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity ∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle ∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH) ΔDAC ≅ ΔDAG . . . HA theorem DC ≅ DG . . . . . . . . . CPCTC ∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC ∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles ∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG) ∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle (∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2) ∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6) This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)
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<u>solution</u>
W=f×d×cos theta
=25N×16m×cos 45°
=400Nm×cos 45°
=282.84J
W=f×d×cos theta
=25N×7m×cos 10°
=175Nm×cos10°
=172.34J
Total workdone=282.84J+172.34J=455.18J