问题1. ABC是一个三角形，D是BC的中点。从D到AB和AC的垂直线相等。证明三角形是等腰的。
解决方案：
Given: D is the mid-point of BC so, BD = DC, ED = FD, and ED ⊥ AB, FD ⊥ AC, so ED = FD
Prove: ΔABC is an isosceles triangle
In ΔBDE and ΔCDF
ED = FD [Given]
BD = DC [D is mid-point]
∠BED = ∠CFD = 90°
By RHS congruence criterion
ΔBDE ≅ ΔCDF
So, now by C.P.C.T
BE = CF … (i)
Now, in △AED and △AFD
ED = FD [Given]
AD = AD [Common]
∠AED = ∠AFD = 90°
By RHS congruence criterion
△AED ≅ △AFD
So, now by C.P.C.T
So, EA = FA … (ii)
Now by adding equation (i) and (ii), we get
BE + EA = CF + FA
AB = AC
So, ΔABC is an isosceles triangle because two sides of the triangle are equal.
Hence proved
问题2。ABC是一个三角形，其中BE和CF分别是与AC和AB边垂直的垂直线。如果BE = CF，则证明ΔABC为等腰。
解决方案：
Given: BE ⊥ AC, CF ⊥ AS, BE = CF.
To prove: ΔABC is isosceles
In ΔBCF and ΔCBE,
∠BFC = CEB = 90° [Given]
BC = CB [Common side]
And CF = BE [Given]
By RHS congruence criterion
ΔBFC ≅ ΔCEB
So, now by C.P.C.T
∠FBC = ∠EBC
∠ABC = ∠ACB
and AC = AB [Because opposite sides to equal angles are equal]
So, ΔABC is isosceles
Hence proved
问题3.在其臂内某个角度内的任何点的垂线是否相等。证明它位于那个角度的等分线上。
解决方案：
Let us consider ∠ABC and BP is an arm within∠ABC
So now draw perpendicular from point P on arm BA and BC, i.e., PN and PM
Prove: BP is the angular bisector of ∠ABC.
In ΔBPM and ΔBPN
∠BMP = ∠BNP = 90° [Given]
MP = NP [Given]
BP = BP [Common side]
So, by RHS congruence criterion
ΔBPM ≅ ΔBPN
So, by C.P.C.T
∠MBP = ∠NBP
and BP is the angular bisector of ∠ABC.
Hence proved
问题4.在图中，AD⊥CD和CB⊥CD。如果AQ = BP且DP = CQ，则证明∠DAQ=∠CBP。
解决方案：
Given that AD ⊥ CD, CB ⊥ CD, AQ = BP and DP = CQ,
Prove:∠DAQ = ∠CBP
We have DP = CQ
So by adding PQ on both sides, we get
DP + PQ = CQ + PQ
DQ = CP … (i)
In ΔDAQ and ΔCBP
We have
∠ADQ = ∠BCP = 90° [Given]
And DQ = PC [From (i)]
So, by RHS congruence criterion
ΔDAQ ≅ ΔCBP
So, by C.P.C.T
∠DAQ = ∠CBP
Hence proved
问题5. ABCD是一个正方形，X和Y分别是AD和BC边上的点，因此AY = BX。证明BY = AX，∠BAY=∠ABX。
解决方案：
In ABCD square,
X and Y are points on sides AD and BC
So, AY = BX.
To prove: BY = AX and ∠BAY = ∠ABX
Now, join Band X, A and Y
So,
∠DAB = ∠CBA = 90° [Given ABCD is a square]
Also, ∠XAB = ∠YAB = 90°
In ΔXAB and ΔYBA
∠XAB = ∠YBA = 90° [given]
AB = BA [Common side]
So, by RHS congruence criterion
ΔXAB ≅ ΔYBA
So, by C.P.C.T
BY = AX
∠BAY = ∠ABX
Hence proved
问题6.以下哪些陈述是正确的(T)，哪些是错误的(F)：
(i)与三角形相等角度相反的一面可能不相等。
(ii)与三角形的等边相反的角度相等
(iii)等边三角形每个角度的量度为60
(iv)如果从一个三角形的一个顶点开始的高度一分为二，则该三角形可能是等腰的。
(v)三角形的两个相等角度的等分线相等。
(vi)如果三角形的垂直角的等分线将底等分，则该三角形可能是等腰的。
(vii)对应于三角形的两个相等边的两个高度不必相等。
(viii)如果直角三角形的任意两个边分别等于另一个直角三角形的两个边，则这两个三角形是全等的。
(ⅸ)两个直角三角形是全等的，如果T他斜边与一个三角形的边分别等于斜边与另一三角形的一侧。
解决方案：
(i) False
(ii) True
(iii) True
(iv) False
(v) True
(vi) False
(vii) False
(viii) False
(ix) True
问题7.填补空白在下面的内容中，使每个以下陈述为真。
(i)与三角形的相等角度相反的边是___
(ii)与三角形的等边相反的角度是___
(iii)在等边三角形中，所有角度均为_
(iv)在ΔABC中，如果∠A=∠C，则AB =
(v)如果三角形ABC的CE和BF高度相等，则AB ___
(vi)在等腰三角形ABC中，AB = AC，如果BD和CE是其高度，则BD为___ CE。
(vii)在直角三角形ABC和DEF中，如果斜边AB = EF而侧面AC = DE，则ΔABC≅Δ___
解决方案：
(i) Equal
(ii) Equal
(iii) Equal
(iv) AB = BC
(v) AB = AC
(vi) BD is equal to CE
(vii) ΔABC ≅ ΔEFD.