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It is known that in △ ABC, ∠ a = 90 °, ab = AC, D is the midpoint of AC, AE ⊥ BD is at e, and extended AE is crossed with BC at F. it is proved that ∠ ADB = ∠ CDF
It is proved that if Ag = 1, then CG = 1, DH = 12, BH = 32, Tan ∠ DBH = 13, GAF = ∠ DBH, GF = 13ag = 13, FH = gh-gf = 12-13 = 16, Tan ∠ FDH = fhdh = 13 ﹤ DBH = ∠ FDH ? ADB = ∠ DBH + ∠ C, ∠ CDF = ∠ FDH + ∠ C
As shown in the figure, in the isosceles right triangle ABC, ∠ ABC = 90 ° and D is the midpoint on the edge of AC. passing through point D, we make de? DF, intersect AB with E, and intersection BC with F. if AE = 4, FC = 3, find EF length
Connect BD,
∵ in the isosceles right triangle ABC, D is the midpoint on the edge of AC,
ν BD ⊥ AC (three wires in one), BD = CD = ad, ∠ abd = 45 °,
∴∠C=45°，
∴∠ABD=∠C，
And ∵ de ∵ DF,
∴∠FDC+∠BDF=∠EDB+∠BDF，
∴∠FDC=∠EDB，
In △ EDB and △ FDC,
A kind of
∠EBD＝∠C
BD＝CD
∠EDB＝∠FDC ，
∴△EDB≌△FDC（ASA），
∴BE=FC=3，
If AB = 7, then BC = 7,
∴BF=4，
In RT △ EBF,
EF2=BE2+BF2=32+42，
∴EF=5．
A: the length of EF is 5
In the isosceles right triangle ABC, the angle ABC is 90, D is on AC, AE is perpendicular to BD, AE is extended to intersect BC at F, angle ADB is equal to angle FDC, and D is proved to be the midpoint of AC
LZ OK, take the midpoint G on DF, and connect AG. ∵ AE ⊥ BC BC ∥ ad →∵ fad = RT ∧ 90 ° → △ fad is RT △ (right triangle) g is the midpoint on the hypotenuse DF,
In the isosceles right triangle ABC, ∠ ABC = 90 ° D is the midpoint of the AC side, passing through point D as De, DF, AB at e, BC at F, if AE = 4, FC = 3, find the EF length Connect BD, ∵ in the isosceles right triangle ABC, D is the midpoint on the edge of AC, ∴BD⊥AC,BD=CD=AD,∠ABD=45°, ∴∠C=45°, And DF, ∴∠FDC=∠EDB, ∴△EDB≌△FDC, ∴BE=FC=3, If AB = 7, then BC = 7, ∴BF=4, In the right triangle EBF, EF^2=BE^2+BF^2=3^2+4^2, ∴EF=5． A: the length of EF is 5 Why BD = CD = ad
Connect BD, ∵ in the isosceles right triangle ABC, D is the middle point on the edge of AC,
In the isosceles right triangle ABC, ∠ ABC = 90 ° D is the midpoint on the edge of AC. passing through point D, de and DF, crossing AB and E, crossing BC at F, if AE = 4, FC = 3
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, ∠ ABC = 90 ° and D is the midpoint on the edge of AC. passing through point D, we make de? DF, intersect AB with E, and intersection BC with F. if AE = 4, FC = 3, find EF length
Connect BD, ∵ in the isosceles right triangle ABC, D is the middle point on the edge of AC, ? BD ⊥ AC (three lines in one), BD = CD = ad, ∵ abd = 45 °, C = 45 °, abd = ∠ C, and ∵ De, DF, ? FDC + ∠ BDF = ∠ EDB + ∠ FDC = ∠ EDB, in △ EDB and △ FDC, ? EBD CB
It is known that in △ ABC, ∠ a = 90 °, ab = AC, D is the midpoint of AC, AE ⊥ BD is at e, and extended AE is crossed with BC at F. it is proved that ∠ ADB = ∠ CDF
It is proved that the vertical lines of BC are made through a and D respectively, and the vertical feet are g and H
Let Ag = 1, then CG = 1, DH = 1
2，BH=3
2，
tan∠DBH=1
3，
And ∠ GAF = ∠ DBH,
∴GF=1
3AG=1
3，
FH=GH-GF=1
2-1
3=1
6，
tan∠FDH=FH
DH=1
Three
∴∠DBH=∠FDH
∵∠ADB=∠DBH+∠C，
∠CDF=∠FDH+∠CDH，
∴∠ADB=∠CDF．
It is known that in △ ABC, ∠ a = 90 °, ab = AC, D is the midpoint of AC, AE ⊥ BD is at e, and extended AE is crossed with BC at F. it is proved that ∠ ADB = ∠ CDF
It is proved that if Ag = 1, then CG = 1, DH = 12, BH = 32, Tan ∠ DBH = 13, GAF = ∠ DBH, GF = 13ag = 13, FH = gh-gf = 12-13 = 16, Tan ∠ FDH = fhdh = 13 ﹤ DBH = ∠ FDH ? ADB = ∠ DBH + ∠ C, ∠ CDF = ∠ FDH + ∠ C
As shown in the figure, in △ ABC, it is known that ∠ a = 90 °, ab = AC, D is a point on AC, AE ⊥ BD is in E, and extended AE is crossed with BC in F. ask: when point d satisfies what conditions, ∠ ADB = ∠ CDF, please explain the reason
When D is the midpoint of AC, ∠ ADB = ∠ CDF. Reason: Ag bisection is made through a ∠ BAC, and BD is intersected with G,
It is known that in △ ABC, ∠ a = 90 °, ab = AC, D is the midpoint of AC, AE ⊥ BD is at e, and extended AE is crossed with BC at F. it is proved that ∠ ADB = ∠ CDF
It is proved that if Ag = 1, then CG = 1, DH = 12, BH = 32, Tan ∠ DBH = 13, GAF = ∠ DBH, GF = 13ag = 13, FH = gh-gf = 12-13 = 16, Tan ∠ FDH = fhdh = 13 ﹤ DBH = ∠ FDH ? ADB = ∠ DBH + ∠ C, ∠ CDF = ∠ FDH + ∠ C
RELATED INFORMATIONS
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- 20. As shown in the figure, ad is the center line of △ ABC, ∠ ADC = 45 ° and BC = 2cm. Fold △ ACD along ad so that point C falls at the position of E, then be=______ cm．