∵AD split ∠BAC? DE⊥AB？ DC⊥AC
∴DE=CD
BC = 24？
CD:DB=3:5
∴CD=24×3/8=9=DE
That is, the distance from point D to AB is 9. Sum up and sublimate? The point on the bisector of an angle is equal to the distance on both sides of the angle. Draw inferences from others? Variant as shown in the figure? ∠ ACB = 90, BD is divided equally ∠ABC is given to AC in D? DE⊥AB in e? ED's extension cable to BC's extension cable is in F, can you verify it? AE=CF
The answer ∵BD split equally ∠ABC? △ADE and △ fcd ∴△ ade △ fcd (ASA) ∴ AE = DE⊥AB,DC⊥BF ∴DE=DC in CFII? Judgment of angular bisector II. Is it known? As shown? CE⊥AB,BD⊥AC,∠B=∠C？ Bf = cf. verification? AF is the bisector of ∠BAC. Ideas? From the known conditions and conclusions to be verified? The judgment theorem of angular bisector should be thought of. Analysis? ∵CE⊥AB,BD⊥AC？ Known? ∴∠ CDF = ∠ BEF = 90 ∠ DFC = ∠ BFE (opposite vertex angle is equal) BF=CF (known) ∴△ DFC △ EFB (AAS) ∴ DF = known? ∴ Point F is on the bisector of ∠BAC (the points with the same distance on both sides of an angle are on the bisector of this angle), that is, the bisector of ∠BAC with AF is summarized and sublimated? Don't leave out the "vertical" condition when applying the bisector theorem and the inverse theorem. If you omit the explanation, you will not realize the necessity of the "vertical" condition in proving the conclusion.
Draw inferences from others? Variant as shown in the figure? It is known that AB=AC, AD=AE, DB and CE intersect at O (1). What if DB ⊥ AC, CE ⊥ AB, D and E are vertical feet? Try to judge the position of O point and the relationship between OE and OD. And prove your conclusion. (2) if d? Didn't e lose her foot? Do you have the same conclusion? And prove your conclusion.
The answer? 1? Ab = ac, ad = AE∴be = CD∫db⊥AC, CE ⊥ AB, ∴∠ BeO = ∠ CDO = 90 in △BEO and △CDO ∴△ BeO. 2? When the feet are vertical, points D and E are not 1? The conclusion still stands? Connect AO in △ABD and △ ace ∴△ Abd △ ace ∴∠ B = ∠ C: AB = AC, AD=AE ∴EB=CD in △BEO and △CDO ∴△ BeO ∴ EO = Do in △AEO. AD⊥BE verification? ∠BAD=∠DAC+∠C
Mind coach: the problem of proving that one angle is equal to the sum of the other two angles? There are generally two ways? 1. Turn two corners into one corner? Prove the equiangle again. 2. Make an angle in the corner? Make it equal to one of these two angles? Then the rest is equal to another angle. Analysis? Do CF⊥BE after c? Is the extension line of the intersection at ∵AD⊥BE? Cf ⊥ be ∴ ad//cf ∴ DAC = ∠ FCA means ∠ FCB = ∠ ACB+∠ DAC in RT △ BCF ∠ FCB = 90-∠ EBC in RT. When adding rails? We should make full use of the known conditions. Draw inferences from others? Variants in quadrilateral ABCD? BC? Ba? AD？ CD？ BD divides equally ∠ABC. Verification? ∠∠?
The answer has passed? Make a perpendicular to the line where AB and BC are? The vertical feet are e, f∶BD bisecting ∠ABC ∴DE=DF and ∶ad = CD ∴△aed△cdf? HL？ ∴∠C=∠DAE and ∠ bad+∠ DAE =180 ∴∠ c+∠ bad =180.
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