As shown in the figure, in the isosceles trapezoid ABCD, it is known that ad ∥ BC, ab = CD, AE ⊥ BC are in E, ∠ B = 60 °, DAC = 45 ° and AC = 6?

As shown in the figure, in the isosceles trapezoid ABCD, it is known that ad ∥ BC, ab = CD, AE ⊥ BC are in E, ∠ B = 60 °, DAC = 45 ° and AC = 6?

For ad \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\adfeis a parallelogram, The perimeter of trapezoidal ABCD is AD + DC + BC + AB = 3-1 + 2 + 2 + 2 + 3-1 = 4 + 23

As shown in the figure, in the isosceles trapezoid ABCD, ad is parallel to BC, ad = AB, BD is vertical to CD, then the angle A=

Angle a = angle ADC = angle ADB + 90
Angle ADB = angle abd
Angle ADB + angle abd + angle a = 180
Angle a + 2 angle ADB = 180
3-angle ADB + 90 = 180
Angle ADB = 30
Angle a = 180-2 * 30 = 120

As shown in the figure, the side length of the square ABCD is 1, P is the midpoint of the side of CD, and the point q is on the line BC (but point B). Let BQ = k, when k=______ The triangle with Q, C and P as vertices is similar to △ ADP

When △ ADP ≌ △ PCQ, k = 0; when △ ADP ≌ △ PCQ, ad: DP = PC: CQ, the side length of ∵ square ABCD is 1, P is the middle point of CD, ad = 1, PD = 0.5, PC = 0.5, CQ = 1-k, that is, 10.5 = 0.51 − K, the solution is k = 0.75

As shown in the figure, in square ABCD, its side length is 1, P is the midpoint of CD, and point q is on the line BC. When BQ is what value, △ ADP is similar to △ QCP?

When ADCP = dpcq, BQ = 34, ∠ d = ∠ C, so △ ADP is similar to △ QCP. When ADCP = DPCP, BQ = 0, △ ADP is similar to △ QCP. Therefore, when BQ = 34 or 0, it can be judged that △ ADP is similar to △ QCP

As shown in the figure, in square ABCD, its side length is 1, P is the midpoint of CD, and point q is on the line BC. When BQ is what value, △ ADP is similar to △ QCP?

When ADCP = dpcq, BQ = 34, ∠ d = ∠ C, so △ ADP is similar to △ QCP. When ADCP = DPCP, BQ = 0, △ ADP is similar to △ QCP. Therefore, when BQ = 34 or 0, it can be judged that △ ADP is similar to △ QCP

In the triangle ABC, ab = AC, ad is perpendicular to point D, point P is on BC, PE is perpendicular to BC, the extension line of Ba is intersected at point E, AC is intersected at point F, 2ad = PE + PF is proved

Certification:
Make GC ⊥ BC cross be extension line to g, make eh ⊥ GC to H
Then ∠ GEH = ∠ B = ∠ FCD
EH=CP
∴Rt△GEH≌Rt△FCP
∴HG=PF
∴PE+PF=CH+HG=CG
The vertical line at the bottom of an isosceles triangle is the center line
The ad is the median of △ BCG
∴2AD=CG=PE+PF

It is known that, as shown in the figure, in △ ABC, ab = AC, D is the point on AC, de ⊥ BC is made through D at e, and the extension line of Ba intersects at F, proving that ad = AF

Certification:
∵DE⊥BC
∴∠FEB=∠DEC=90º
∵AB=AC
∴∠B=∠C
∵∠B+∠F=90º,∠C+∠CDE=90º
∴∠F=∠CDE=∠FDA
∴AD=AF

As shown in the figure, in △ ABC, ab = AC, e is on AC, and the extension line of ad = AE, de intersects BC at point F

It is proved that: as shown in the figure, am ⊥ BC is made in M through a, ∵ AB = AC, ∵ BAC = 2 ⊥ BAM, ∵ ad = AE, ∵ d = ∠ AED, ∵ BAC = 2 ⊥ BAM = 2 ⊥ D, ∵ BAM = D, ∵ DF ∥ am, ≁ am ⊥ BC, ≁ DF ⊥ BC

Known: as shown in the figure, in △ ABC, ab = AC, BC = BD, ad = de = EB, then the degree of ∠ A is ()
A. 30°B. 36°C. 45°D. 50°

In the case of EBD (EBD = x °, and the "be = De,, \\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\1

As shown in the figure, in △ ABC, ∠ BAC = 108 ゜, ab = AC, BD bisects ∠ ABC, intersects AC with D, and proves: BC = CD + ab

In △ abd and △ EBD, in △ abd and △ EBD, in △ abd and △ EBD, in △ abd and △ EBD, ab = EB, ab = EB, abd = abd = ebdbd = BD, be = Ba = Ba, connect De, De, \8780; ≌ EBD (SAS), connect de, and W BD 87; BD, bisbisbisbisbisbisbisbisbisbisbisbisbisbisbisdivide \\\\A BD