It is known that: as shown in the figure, the circle O with the diameter of AB on one side of the equilateral triangle ABC intersects with the sides AC and BC at points D and e respectively, DF ⊥ BC is made through point D, and F is the perpendicular foot (1) Proof: DF is tangent of circle O (2) If the side length of equilateral triangle ABC is 4, find the length of DF (3) Finding the area of the shadow in a graph The first question is not, the shadow area is DFE

It is known that: as shown in the figure, the circle O with the diameter of AB on one side of the equilateral triangle ABC intersects with the sides AC and BC at points D and e respectively, DF ⊥ BC is made through point D, and F is the perpendicular foot (1) Proof: DF is tangent of circle O (2) If the side length of equilateral triangle ABC is 4, find the length of DF (3) Finding the area of the shadow in a graph The first question is not, the shadow area is DFE

2)
AD=DC=AO=2=BC/2
DF=CD*sinC=√3
3）
CF=EF=1/2CD=1
S triangle def = 1 / 2 * DF * EF = √ 3 / 2

As shown in the figure, in the triangle ABC, ad ⊥ BC, EA ⊥ Ca, ∠ BAE = ∠ C, try to explain: Ed / ad = be / ab

AD⊥BC,EA⊥CA,
So: △ EDA ∽ eac
So: Ed / ad = AE / AC
Because: ∠ BAE = ∠ C, ∠ A is common
So: △ bea ∽ bac
So: be / AB = AE / AC
So: Ed / ad = be / AB

In the triangle ABC, CE ` BD are the heights on the sides of AB ` AC respectively. It is proved that the vertical bisector of de passes through the midpoint m of BC
There should be a specific process

In the triangle ABC, CE ` BD is the height of AB ` AC side respectively. It is proved that the vertical bisector of de passes through the midpoint m of BC. It is proved that connecting DM, EM ∵ CE ` BD is the height of AB ` AC side respectively. M is the midpoint of BC. DM = BC / 2, EM = BC / 2

As shown in the figure, in △ ABC, BD and CE are the midlines on edge AC and ab respectively, and BD = CE, ab = AC

D and E are respectively AC, the midpoint of ab \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ab = AC

It is known that, as shown in the figure be, CF is the height of the edges AC and ab of △ ABC. Intercept BP = AC on be and CQ = AB on the extension line of CF. verify: AP ⊥ aq

It is proved that: ∵ CF ⊥ AB, be ⊥ AC, ∩ AEB = ≌ AFC = 90 °, ∩ Abe = ≌ ACQ = 90 ° - BAC. ∵ BP = AC, CQ = AB, in △ APB and △ QAC, BP = AC ⊥ Abe = ≌ acqcq = AB, ≌ APB ≌ QAC (SAS). ∩ BAP = ≌ CQA. ∩ CQA + ∩ qaf = 90 ° and ≁ BAP + ≌ qaf = 90 °. That is AP ⊥ AQ

As shown in the figure, be and CF are respectively the heights of obtuse angle △ ABC (angle a > 90 °). Intercept BP = AC on be and CQ = AB on the extension line of CF to connect AP and aq
(1) Is AP equal to AQ? Explain why
(2) Judge the position relationship between AP and AQ, and prove your judgment

1. It is proved that: be ⊥ CE, CF ⊥ BF ≁ Abe ≁ BAE = 90, ≁ ACF ≌ CAF = 90 ≌ Abe ≌ BAE ≁ caf ≁ BAE ≁ caf ≁ Abe ≁ BP = AC, CQ = ab ≌ ABP ≌ QCA (SAS) ≌ AP = aq2, AP ⊥ aq

As shown in the figure, in △ ABC, AB > AC, be and CF are all high of △ ABC, P is a point on be and BP = AC, q is a point on CF extension line and CQ = AB, connect AP, AQ and QP, and verify: (1) AP = AQ; (2) AP ⊥ aq

The following: (1) all of the be and CF are the high of △ ABC, and the \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\faq = 90 °, i.e. ∠ PAQ = 90 ° ，∴AP⊥AQ．

As shown in the figure, given that be and CF are high on both sides of the triangle ABC, intercept BP = AC on be and CQ = AB on the extension line of CF. is pa perpendicular to AQ?

Angle ABP = 90 degrees - angle BAC = angle ACQ,
CQ=AB,BP=AC,
Triangle ABP and QCA are congruent,
Angle q = angle BAP,
AP, CQ, O,
Angle BAP + angle AOQ = 90 degrees,
Angle Q + angle AOQ = 90 degrees,
PA perpendicular to AQ

Answer one question: we know that be and CF are the heights of triangle ABC. If we intercept BP = AC on be and CQ = AB on the extension line of CF, is pa perpendicular to AQ?
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Because be and CF are high, AEB and AFC are right triangles, so angle ABP = angle ACF
Because AB = CQ, BP = AC, so triangle ABP is equal to triangle QCA, so angle QAC = angle APB
Angle QAP = angle QAC - angle PAC, and angle PAC + angle bea = angle APB, so angle QAP = angle bea = 90 degrees
PA perpendicular to AQ

It is known that, as shown in the figure be, CF is the height of the edges AC and ab of △ ABC. Intercept BP = AC on be and CQ = AB on the extension line of CF. verify: AP ⊥ aq

It is proved that: ∵ CF ⊥ AB, be ⊥ AC, ∩ AEB = ≌ AFC = 90 °, ∩ Abe = ≌ ACQ = 90 ° - BAC. ∵ BP = AC, CQ = AB, in △ APB and △ QAC, BP = AC ⊥ Abe = ≌ acqcq = AB, ≌ APB ≌ QAC (SAS). ∩ BAP = ≌ CQA. ∩ CQA + ∩ qaf = 90 ° and ≁ BAP + ≌ qaf = 90 °. That is AP ⊥ AQ