BD is the bisector of angle ABC, De is perpendicular to AB and E, DF is perpendicular to BC and F, area of triangle ABC = 36, ab = 18, BC = 12, find the length of De and DF
Triangle BDE and triangle BDF are congruent triangles, then de = DF
And the area of triangle ABC is equal to the sum of the area of triangle abd and triangle ACD,
Triangle abd area = (1 / 2) de * AB,
Triangle ACD area = (1 / 2) DF * BC
That is, 36 = (1 / 2) * 18 * de + (1 / 2) * 12 * DF = 9de + 6df = 15de,
De = DF = 36 / 15
In the triangle ABC, ab = AC, D is a point on AB, ad = 2 | 3AB, DF | BC, e is the midpoint of BD, if EF, AC, BC = 6, the area of quadrilateral dbcf is obtained
Take FC midpoint m to connect em
∴DE=BE FM=CM
The EM is the median line of trapezoidal dfcb
∴EM=(4+6)÷2=5
And DF ‖ EM ‖ BC
Do FK ⊥ BC to FM at G-spot
∴∠FKE=90°
From the known ∠ EFC = 90 °
∴∠EFG+∠CFK=∠EFG+∠FEM
∴∠CFK=∠FEM
∵∠EFC=∠FKC=90°
∴△EFM∽△FKC
∵ ad = 2Ab / 3
Let de = be = X
Then ad = 4x
∵DF∥BC
∴△ADF∽△ABC
AD:AB=DF:BC=4X:6X=2:3
∵ BC = 6, so DF = 4
Do vertical BC
∴△DBO≌△FCK
∴CK=(6-4)÷2=1 CF=DB=2X
∵△EFM∽△FKC
∴FC:CK=EM:FM
∴2X:1=5:X
The solution is x = √ 10 / 2 (root of two 10)
Ψ FC = √ 10 (root 10)
The Pythagorean theorem gives high FK = 3
∴S=(DF+BC)×FK÷2
=(4+6)×3÷2
=15
In the triangle ABC, point D is the point on the edge of BC, ad = CD, f is the midpoint of AC, de bisects the triangle ADB, intersects AB at point E, and proves that De is perpendicular to DF
In triangle ADF and triangle CDF, ad = CD, AF = CF, DF are common edges, so triangle ADF and triangle CDF are fully equal angle ADF = angle CDF = angle ADC / 2DE bisecting angle ADB, so angle ade = angle BDE = angle ADB / 2, so angle ADF + angle ade = (angle ADC / 2) + (angle ADB / 2) = (angle ADC + angle ADB) / 2 = 180 / 2 = 90 degrees, so De is vertical D
In Δ ABC, ab = 8, AC = 6, ad is its angular bisector, de ⊥ AB, DF ⊥ AC, the perpendicular feet are e and f respectively. If the area of Δ abd is 12, calculate the area of Δ ACD
DE=SΔABDX2/AB=3
Because ad is an angular bisector, de ⊥ AB, DF ⊥ AC,
So de = DF [the distance from one point of the bisector to both sides of the angle, etc.]
So s Δ ACD = dfxac / 2 = 9