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).)
We know the y intercept is -6 and we know the slope is positive 4 (rise over run) so the equation is y=4x-6 if we plug in a shaded point (I would choose 0,0 for convenience reasons) 0=-6 since -6 is less than 0, the expression would be y≥4x-6
Answer:
D) 3
Step-by-step explanation:
It is halfway the length of Y and W. 6-3= 3
Note:
Pls notify me if my answer is incorrect for the other users that will see this response. Thank you.
<em>-kiniwih426</em>
Answer:
BD = 3 units
Step-by-step explanation:
Since, AD is an angle bisector of ∠BAC,
m∠BAD = m∠CAD = 20°
CD = 3 units
In ΔACD and ΔABD,
m∠BAD = m∠CAD = 20° (Given)
AD ≅ AD [Reflexive property]
Therefore, by H-A property of congruence both the triangles will be congruent.
And by CPCTC,
CD ≅ BD = 3 units
