In the isosceles right triangle ABC, the angle ABC = 90 degrees D is the midpoint on the edge of AC, where de is perpendicular to DF and intersects AB at If AE = 4, FD = 3, find the length of EF

Connect BD, ∵ in the isosceles right triangle ABC, D is the middle point on the edge of AC,

As shown in the figure, in the isosceles right triangle ABC, the angle ABC = 90 degrees, D is the midpoint of AC side, and De is made through point D, perpendicular to DF, and intersecting AB at point E If AE = 4, FC = 3, find the length of BF, be

Connect ad, ∵ Ba = BC, D is the midpoint of AC,
∴BD⊥AC,
∴∠CDF+BDF=90°,
∵DE⊥DF,∴∠BDE+∠BDF=90°,
∴∠CDF=∠BDE,
∵∠ABC=90°,
∴∠DBE=∠C=45°,BD=1/2AC=CD,
∴ΔBDE≌ΔCDF(ASA),
∴BE=CF=3,
∴BF=AE=4.

The triangle ABC is an isosceles right triangle, ab = AC, D is the midpoint on the hypotenuse BC, e and F are the points on the sides of AB and AC respectively, and De is vertical DF If be = 8cm, CF = 6cm, calculate the area of triangular def Hurry up, those who will come, Sweat, if it's that easy, do I need to ask questions-

Connecting ad
Ad = < CD ＜ 45 °)
So CDF is all equal to DAE
So AE = CF = 6
Similarly, AF = 8
So EF = 10
Because of congruence, DF = De
That is, Fed is isosceles right triangle
The area is 25
The process is not complete. You can ask me if you don't understand

The triangle ABC is an isosceles right triangle, ab = AC, D is the midpoint of the hypotenuse BC, e f is the point on the edge of AB AC, and De is vertical DF, if be = 12, CF = 5 Find the length of the segment ef

The key is to prove that the triangle DCF is equal to the triangle DAE
Angle c = angle DAB is 45 degrees
CD=DA
Angle CDF = angle ade and angle FDA complementary
AE = CF = 5 is obtained
af=12
So EF = 13

As shown in the figure, △ ABC is an isosceles right triangle, ab = AC, D is the midpoint on the hypotenuse BC, e, f are the points on the sides of AB and AC respectively, and de ⊥ DF, if be = 8, CF = 6, then s ⊥ def=______ .

As shown in the figure, the crossing point D is DG ⊥ AB in G, DH ⊥ AC in H, ∵ △ ABC is an isosceles right triangle, D is the middle point on the hypotenuse BC,  DG = BG = DH = ch, ∵ GDH = 90 °, ∵ EDG + ∠ EFH = 90 °, ∵ de ⊥ DF,  FDH + ∠ EDH = 90 ° and  EDG = ∠ FDH

As shown in the figure, △ ABC is a right triangle, ∠ cab = 90 °, D is the midpoint on the hypotenuse BC, e and F are the points on AB and AC respectively, and de ⊥ DF (1) If AB = AC, be = 12, CF = 5, calculate the area of △ def (2) It was proved that be 2 + CF 2 = EF 2

(1) Then:  CDF ≌ ≌ ≌≌≌≌≌≌≌≌≌≌≌≌≌≌≌≌ BDG  DBG ≌ DBG ≌ DBG ≌ DBG ? C ᙱ BG = CF = 5 ᙨ eg = 5 ᙨ eg ᙨ eg ≌ ADF ≌ ADF ? CDF CD, ∠ DAE = C

As shown in the figure, D is the midpoint of AB side in △ ABC, △ ace and △ BCF are isosceles right triangle with AC and BC as oblique sides, connecting de and DF Confirmation: de = DF

It is proved that: take the midpoint m and N of AC and BC respectively, connect MD and Nd, and then connect EM and FN, ∵ D is the midpoint of AB, ∵ AEC = 90 °, BFC = 90 °, EM = 12ac, FN = 12bc, ∵ D is the midpoint of AB edge in △ ABC, ? DN is the median line of △ ABC. ? DN = 12ac, ? EM = DN = 12ac, FN = MD = 12bc, ? DN

It is known that in isosceles RT △ ABC, ∠ a = 90 ° D is the midpoint of BC, e and F are points on AB and AC respectively, and EA = CF is satisfied. Verification: de = DF

Proof: connect ad, as shown in the figure,
∵ △ ABC is an isosceles right triangle, D is the midpoint of BC,
ν ad = DC, ad bisection ∠ BAC, ∠ C = 45 °,
∴∠EAD=∠C=45°，
In △ ade and △ CDF
EA＝CF
∠EAD＝∠C
AD＝CD ，
∴△ADE≌△CDF，
∴DE=DF．

It is known that in isosceles RT △ ABC, ∠ a = 90 ° D is the midpoint of BC, e and F are points on AB and AC respectively, and EA = CF is satisfied. Verification: de = DF

In the isosceles right triangle ABC, the angle ABC = 90 degrees D is the midpoint on the edge of AC, where de is perpendicular to DF and intersects AB at If AE = 4, FD = 3, find the length of EF