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 and ac . point g ∈ ab so that dg ⊥ ab and h ∈ ab so that eh ⊥ ab .prove m∠hcg = 45° (with statement reason preferably).

This is a little long, but it gets you there.

ΔEBH ≅ ΔEBC . . . . HA theoremEH ≅ EC . . . . . . . . . CPCTC∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHCΔAHE ~ 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 theoremDC ≅ 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).)

The answer is 270 cars.you can do this by making a proportional relationship. for instance, 200 is the whole we are using, so on a “number line” 200 would be over 100%. now, since he sells 35% more, add 35% to 100% to get 135% on the other side of the “number line” on the bottom half, leaving the top half empty.cross multiply 200 by 135 then divide by 100 to get your answer. i hope this , if it did, , if you don’t mind, giving me brainiest! 200/100% } 200 • 135 /100 ? /135%