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As shown in the figure, △ ABC is an isosceles right triangle, ∠ a = 90 ° and points P and Q are moving points on AB and AC respectively, and satisfy BP = AQ, D is the midpoint of BC (1) It is proved that △ PDQ is an isosceles right triangle; (2) When the point P moves to where, the quadrilateral apdq is square, and explain the reason
(1) Proof: connect ad
∵ △ ABC is an isosceles right triangle and D is the midpoint of BC
∴AD⊥BC，AD=BD=DC，∠DAQ=∠B，
In △ BPD and △ aqd,
BD＝AD
∠DBP＝∠DAQ
BP＝AQ ，
∴△BPD≌△AQD（SAS），
∴PD=QD，∠ADQ=∠BDP，
∵∠BDP+∠ADP=90°
Ψ ADP + ∠ ADQ = 90 °, i.e. ∠ PDQ = 90 °,
The △ PDQ is an isosceles right triangle;
(2) The reason for moving to the midpoint of the square is as follows:
∵ BAC = 90 °, ab = AC, D is the midpoint of BC,
∴AD⊥BC，AD=BD=DC，∠B=∠C=45°，
ν Δ abd is an isosceles right triangle,
When p is the midpoint of AB, DP ⊥ AB, i.e., ≁ APD = 90 °,
And ? a = 90 °, PDQ = 90 °,
The quadrilateral apdq is rectangular,
And ∵ DP = AP = 1
2AB，
The rectangle apdq is a square (a rectangle with equal adjacent sides is a square)
As shown in the figure, △ ABC is an isosceles right triangle, ∠ a = 90 ° and points P and Q are moving points on AB and AC respectively, and satisfy BP = AQ, D is the midpoint of BC (1) It is proved that △ PDQ is an isosceles right triangle; (2) When the point P moves to where, the quadrilateral apdq is square, and explain the reason
(1) It is proved that: connecting ad ∵ ABC is an isosceles right triangle, D is the midpoint of BC ∵ ad ⊥ BC, ad = BD = DC, ∵ DAQ = ∠ B. in △ BPD and △ aqd, BD = ad ∵ DBP ≌ ≌ △ aqd (SAS), PD = QD, ∵ ADQ = ≌ ≌≌△ aqd (SAS), PD = QD, ∠ ADQ = ∠ BDP, ? BDP + ∠ ADP = 90 °
As shown in the figure, the triangle ABC is an isosceles right triangle, the angle a = 90 ° and the points P and Q are the moving points on AB and AC respectively, and satisfy BP = AQ, and point D is the midpoint of BC
1. Connect ad
BP=AQ ∠QAD=∠B=45 AD=BD
△BPD≌△AQD PD=QD
∠PDB=∠QDA ∠QDP=∠AQD+∠ADP=∠PDB+∠ADP=∠ADB=90
Therefore: the triangle PDQ is an isosceles right triangle
When P and Q are the midpoint of AB and AC respectively, the quadrilateral apdq is a square
If AP = BP ad = BD, then PD ⊥ ab ∠ APD = 90
∠PAD=45 AP=PD
Similarly, ∠ aqd = 90, AQ = QD
BP=AQ BP=AB
AP=PD=AQ=QD ∠AQD=90 ∠APD=90 ∠QAP=90 ∠QDP=90
Therefore: the quadrilateral apdq is a square
As shown in the figure, RT △ ABC ≌ RT △ ade, ∠ a = 90 °, BC and de intersect at point P. if AC = 6, ab = 8, then the distance from point P to edge AB is______ .
Pass P as PM ⊥ AB to m, PN ⊥ ad to N, and connect AP,
∵Rt△ABC≌Rt△ADE，∠A=90°，AC=6，AB=8，
∴∠B=∠D，AD=AB=8，AC=AE=6，
∴BE=CD=2，
∵ in ᙽ BEP and ∵ DCP,
∠B＝∠D
∠BPE＝∠DPC
BE＝CD
∴△BEP≌△DCP（AAS），
∴PB=PD，
∵PM⊥AB，PN⊥AD，
∴∠BMP=∠DNP=90°，
In △ BMP and △ DNP
∠B＝∠D
∠BMP＝∠DNP
BP＝DP
∴△BMP≌△DNP，
∴PM=PN，
∵S△ABC=S△BAP+S△CAP，
∴1
2×8×6=1
2×8×PM+1
2×6×PN，
∴PM=PN=24
7，
So the answer is: 24
7．
As shown in the figure, in RT △ ABC, ∠ a = 90 °, ab = 6cm, AC = 8cm. Take the point P on the hypotenuse BC which is 6cm away from point B as the center, rotate the triangle 90 ° anticlockwise to △ def, then the area of the overlapping part of the two triangles before and after the rotation is______ cm2．
Make PM ⊥ AC in M, PN ⊥ DF in N, as shown in the figure, ∵ take the point P on the oblique side BC 6cm away from point B as the center, rotate the triangle by 90 ° to △ def, kph = 90 °, KGH = 90 °, MPN = 90 °, ∵ KPN = ∠ mph, ∵ PC = PF, ≌ C = ≌ F, ≌ RT △ PFN
As shown in the figure, △ ABC is a right triangle, ∠ ACB = 90 ° and ∠ 1 = ∠ B. if AC = 8, BC = 6, find the length of CD Answer by 6:00 a.m. on April 29 The picture needs imagination
∠A+∠B=90
∠1=∠B
So ∠ 1 + ∠ a = 90
ν CD vertical ab
CD*AB=AC*BC
CD=4.8
As shown in Fig. 1, it is known that △ ABC is an isosceles right triangle, ∠ BAC = 90 ° and point D is the midpoint of BC As shown in Fig (1) Try to guess the quantitative relationship between line BG and AE. Please write your conclusion directly; (2) After the square defg is rotated anticlockwise around point d (the rotation angle is greater than 0 ° and less than or equal to 360 °), as shown in Fig. 2, whether the conclusion in (1) is still valid through observation or measurement? If so, please prove it; if not, please explain the reason; (3) If BC = de = 2, during the rotation of (2), find the maximum and minimum values of AE length
(1) BG = AE, easy to get BD = Da, Gd = Da, ∠ GDB = ∠ EDA; therefore, RT △ BDG ≌ RT △ ade can be obtained; therefore BG = AE; (2) it is true that connecting ad, ∵ RT △ BAC, D is the middle point of the hypotenuse BC,
It is known that △ ABC is an isosceles right triangle, ∠ BAC = 90 ° and point D is the midpoint of BC. Make square defg and connect AE. If BC = de = 2, rotate square defg anticlockwise around point D. in the process of rotation, when AE is the maximum value, calculate the value of AF
Radical 13+++++++++++++++++++++++++++++++++++
8. As shown in the figure, △ ABC and △ DEA are two congruent isosceles right triangles, ∠ BAC = ∠ d = 90 ° BC intersects AD and AE at points F and g respectively. Answer the following questions: (1) How many triangles are there in the picture? Please show them (2) What pairs of similar triangles are there? Please show them and explain why
1. There are seven triangles; they are △ ABC, △ ABF, △ AFG, △ AGC, △ ABG, △ AFC, △ ade
2. There are two pairs of similar triangles, which are △ Abf ∽ △ AGC, △ ABG ∽ △ AFC
As shown in the figure: in the right triangle ABC, the angle c = 90 ° D is the midpoint of BC, de ⊥ AB and E, tanb = 1 / 2, AE = 7, find the length of de
I can see it best in this way
Let de be y
∵ tanB=½
﹣ EB = 2DE = 2Y, and AC = half CB
According to Pythagorean theorem, DB = y √ 5
∵ D is the midpoint of BC
∴ AC=DB=y√5
In △ ACB
According to Pythagorean theorem, it is found that AB 2 = AC 2 + CB 2
Namely:
(7+2y)²=(y√5)²+(2y√5)²
(7+2y)²=25y²
7+2y=5y
7=3y
It's 3 / y
The length of De is 7 / 3
RELATED INFORMATIONS
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