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As shown in the figure, the chord AB and radius OC of ⊙ o are extended and intersected at point D, BD = OA. If ∠ AOC = 105 °, then ∠ D=______ Degree
Connect ob,
∵BD=OA，OA=OB
So △ AOB and △ BOD are isosceles triangles,
Let ∠ d = x degree, then ∠ oba = 2x °,
Because ob = OA,
So ∠ a = 2x °,
In △ AOB, 2x + 2x + (105-x) = 180,
The solution is x = 25,
That is ∠ d = 25 °
As shown in the figure, it is known that AB is the chord of ⊙ o, the radius OA = 20cm, ∠ AOB = 120 ° and the area of ⊙ AOB is obtained
OC ⊥ AB is set at C through point O, as shown in the following figure:
∴∠AOC=1
2∠AOB=60°，AC=BC=1
2AB，
In RT △ AOC, ∠ a = 30 °
∴OC=1
2OA=10cm，
AC=
OA2−OC2=
202−102=10
3（cm），
∴AB=2AC=20
3cm
The area of △ AOB = 1
2AB•OC=1
2×20
3×10=100
3（cm2）．
In the triangle ABC, D is a point on the edge of BC, e is the midpoint of AD. passing through point a is the parallel line of BC, the extension line of be is at F, and AF = DC, connecting BF 1) Verification: D is the midpoint of BC 2) If AB = AC, try to guess the shape of the quadrilateral adcf and prove your conclusion
(1) because of AF / / BC
So AE / ed = AF / BD must choose me!
Because e is the mid point of AD,
So AE = ed
So AF = BD
Because AF = DC
So BD = DC,
D is the midpoint of BC
(2) Rectangle
Because AB = AC,
And D is the midpoint of BC
So ad vertical BC
Because AF / / is equal to DC
therefore
The quad adcf should be rectangular
As shown in the figure, it is known that in △ ABC, ab = AC, point D is the midpoint of BC edge, be is the height of AC edge, BF is parallel to AE and BF = AE, connecting DF, De Verification: (1) AED = ∠ FBD （2）ED⊥DF
(1) Proof: because be is the height on the edge of AC
So the angle BEC = angle AEC = 90 degrees
So the triangle BEC is a right triangle
Because D is the midpoint on the BC side
So ad and de are the midlines of the triangle ABC and the right triangle BEC respectively
So de = BD
So angle DBE = angle DEB
Because BF is parallel to AE and BF = AE
So the quadrilateral afbe is a parallelogram
So the quadrilateral afbe is a rectangle
So the angle EBF is 90 degrees
Because angle AED = angle AEB + angle DEB = 90 + angle DEB
Angle FBD = angle EBF + angle DBE = 90 + angle DBE
So angle AED = angle FBD
(2) Proof: because BF = AE
BD = de (certified)
Angle FBD = angle AED (proved)
So triangle FBD and triangle AED congruence (SAS)
So angle BDE = angle ade
Because AB = AC
So the triangle ABC is an isosceles triangle
Because ad is the center line of the triangle ABC
So ad is the perpendicular of the isosceles triangle ABC
So angle ADB = angle ADF + angle BDF = 90 degrees
Because angle EDF = angle ADF + angle ade
So the angle EDF is 90 degrees
So Ed vertical DF
As shown in the figure, in △ ABC, D, e and F are the points on AB, AC and BC respectively, de = 3, BF = 9 / 2, AD / AB = AE / AC = 2 / 5, and verify DF ‖ AC
∵AD/AB=AE/AC
∴DE‖BC
∴△ADE∽△ABC
∴DE/BC=AD/AB=2/5
DE/(BF+FC)=2/5
3/（9/2+FC）=2/5
FC=3 BC=15/2
∴BF/BC=9/2/15/9=3/5
Ad / AB = 2 / 5
∴BD/AB=3/5
∴BD/AB=BF/FC
∴DF‖AC
It is known that, as shown in the figure, BD is the bisector of ∠ ABC, ab = BC, point P is on BD, PM ⊥ ad, PN ⊥ CD, and the perpendicular feet are m and N respectively
It is proved that: in △ abd and △ CBD, ab = BC (known), ∠ abd = ∠ CBD (property of bisector), BD = BD (common side), abd ≌ △ CBD (SAS), ∵ ADB = ∠ CDB (the corresponding angles of congruent triangles are equal); ∵ PM ⊥ ad, PN ⊥ CD, ∵ PMD = ∠ PND = 90 °; and ∵ PD = PD (...)
It is known that, as shown in the figure, BD is the bisector of ∠ ABC, ab = BC, point P is on BD, PM ⊥ ad, PN ⊥ CD, and the perpendicular feet are m and N respectively
It is proved that: in △ abd and △ CBD, ab = BC (known), ∠ abd = ∠ CBD (property of bisector), BD = BD (common side), abd ≌ △ CBD (SAS), ∵ ADB = ∠ CDB (the corresponding angles of congruent triangles are equal); ∵ PM ⊥ ad, PN ⊥ CD, ∵ PMD = ∠ PND = 90 °; and ∵ PD = PD (...)
It is known that, as shown in the figure, BD is the bisector of ∠ ABC, ab = BC, point P is on BD, PM ⊥ ad, PN ⊥ CD, and the perpendicular feet are m and N respectively
It is proved that: in △ abd and △ CBD, ab = BC (known), ∠ abd = ∠ CBD (property of bisector), BD = BD (common side), abd ≌ △ CBD (SAS), ∵ ADB = ∠ CDB (the corresponding angles of congruent triangles are equal); ∵ PM ⊥ ad, PN ⊥ CD, ∵ PMD = ∠ PND = 90 °; and ∵ PD = PD (...)
As shown in the figure, BD is the bisector of angle ABC, point P is on BD, PM is perpendicular to ad, PN is perpendicular to CD, points m and N are respectively perpendicular to foot, and PM = PN, verify AB = BC
It is proved that because PM is perpendicular to AD and PN is perpendicular to CD, angle PMD is equal to angle, PND is equal to 90 degrees, PD is equal to PD, PM is equal to PN, so triangle PMD is equal to triangle PND, so angle MDP is equal to angle NDP, because BD is equal to BD, angle abd is equal to angle CBD, so AB is equal to BC
As shown in the figure, D is the midpoint on BC side of triangle ABC, De is perpendicular to AC, DF is perpendicular to AB, and perpendicular feet are points E and f respectively. If BF = CE, then triangle ABC is isosceles
prove:
∵ D is the midpoint of the edge BC of ᙽ ABC
∴BD＝DC
∵DE⊥AC DF⊥AB
∴∠DFB＝∠DEC
And ∵ BF = CE
∴△BDF≡△CDE
∴∠FBD＝∠DCE
The △ ABC is an isosceles triangle
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