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

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