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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）．
As shown in the figure: in △ ABC, ∠ a = 90 °, ab = AC, D is the midpoint of BC, AE = BF, verification (1): de = DF; (2): de ⊥ DF There are pictures There are pictures only can not be sent. To picture + Q 1825157166
It is proved that: 1) if the connected ad. triangle ABC is an isosceles right triangle, then ad ⊥ BC, ad = BD = BC / 2, ∠ DAE = 45 ° = ∠ B
If AE = BF, then ⊿ DAE ≌ Δ DBF (SAS), de = DF
2) ⊿ DAE ≌ Δ DBF (proved), then ≌ ade = ∠ DBF
Therefore: ∠ ade + ∠ ADF = ∠ DBF + ∠ ADF = 90 ° therefore, de ⊥ DF
As shown in the figure, in △ ABC, ∠ ACB = 90 ° AC = BC, point D is the midpoint of AB, AE = CF
It is proved that: as shown in the figure, the connection CD. ∵ BC = AC, ∵ BCA = 90 °, ∵ ABC is an isosceles right triangle, ∵ D is the midpoint of AB, ∵ BD = CD, CD bisection ∠ BCA, CD ⊥ ab. ∵ a + ∵ ACD = ∠ ACD + ∠ FCD = 90 ° and a = ∠ FCD
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 angular bisector),
BD = BD (common side),
∴△ABD≌△CBD（SAS），
Ψ ADB = ∠ CDB (the corresponding angles of congruent triangles are equal);
∵PM⊥AD，PN⊥CD，
∴∠PMD=∠PND=90°；
And ∵ PD = PD (common side),
∴△PMD≌△PND（AAS），
ν PM = PN (the corresponding sides of an congruent triangle are equal)
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 (...)
BD is the bisector of angle ABC, Ba is equal to BC, point P is on BD, and PM is vertical to ad, PM is vertical to CD, and PM is equal to PN Because I'm new,
AB=BC BD=BD ∠ABD=∠CBD
∴ △ABD≌△CBD
∴ ∠ADB=∠CDB
∴ ∠MDP=∠NDP
∠PMD=∠PND=90°
PD=PD
∴ △PDM≌△PDN
∴ PM=PN
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It is known that D is the midpoint on the BC side of the triangle, De is perpendicular to AC, DF is perpendicular to AB, perpendicular is e, F, and BF = CE
In right triangle BDF and CDE
BD=DC
BF=CE
So the triangle BDF is all equal to CDE
So angle B = angle C
So the triangle ABC is isosceles triangle
D is the midpoint on BC side of triangle ABC, De is vertical AC, DF is vertical AB, and perpendicular foot is point E and F. if BF = CE, then triangle ABC is isosceles triangle. Please explain the reason
Triangle BFD, congruent triangle Dec, HL, so angle B equals angle c, so
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