Axisymmetric geometry problems are advanced. Grade 2 Axisymmetric Geometry Problem Solving!

After D, DM is parallel to BC and AB is connected to M because in ABC, ACB=90, BAC=30 and ABEs and ACD are regular triangles.

Therefore, M can be proved to be the midpoint of AB.

Therefore, EM is perpendicular to AB and EM is parallel to AD

Therefore, the quadrilateral AEMD is a parallelogram.

So ef=fd

The process is relatively simple, and I don't understand it and ask again.

Make the perpendicular line of ab from point e and cross ab to g

In ADF, there is DAF= BAC+ CAD=90 and AD=AC= 3AB

In gef, there is egf=90 and ge= 3be= 3ab so there is. 1）∠afd=∠gfe

2）∠daf=∠egf

3）ad=ge

So ADF is congruent with GEF.

So df=ef

The question stem is unclear and cannot be answered.

ab=ac∠b=∠acb

eb=ed b= Extinction edb

acb=∠edb

ef‖acae=be

be=deab=ae+be=ac

ed=df, diff ef=ed+df

The quadrilateral aefc is a flat virtual macro calling quadrilateral.

a=∠f<>

Connect od, because the point e is symmetrical with the point o with respect to the straight line bc, so oed= doc

Because of BC AO, AOB = Lift OBC = 35° and because the point A and point D are symmetrical with respect to the straight line OB, so Lao Bi AOB = DOB = 35°

So oed= doc 90° 2 30° 30°

If there is a problem with the condition you give, it should be ac=bc, and abc is an isosceles right-angled triangle proof: connect cdBecause D is the midpoint of AB, CD=AD and CD AB=AE, the angle A= DCB=45°

So aed afd

So ade= cdf

So edf= edc+ cdf= edc+ ade=90°, i.e. de df

Solution: Connect cd because ae=cfcd=cd a= dcf, so aed is all equal to dcf, so ade= cdf, and because ade+ edc=90°, cdf+ edc=90°, i.e., edf=90°, so de df.

Connecting the CDX three lines and one gives CD vertical ab ad=1 2ab The middle line on the hypotenuse = half of the write edge, i.e., ad=cd, it can be proved that AED is all equal to DCF, so cdf=ade, so EDF=90, i.e., de df

Let the grassland be the point A, and the symmetrical point of A about the river is the point B, connect Pb, and intersect the river at point C. Then the herdsmen start from P and drive the horses to point C to drink water, and then graze at point A and come back along the AP for the shortest distance.

Since ac and bc are symmetrical with respect to the river, then ac=bc, and the straight line pb between p and b is the shortest, so pc+ca is the smallest.

The riverside should be in the same straight line as the meadow, and the meadow should be between the horse camp and the river.

Triangle. Utilizes the shortest point-to-point distance segment.

<> Equivalence BEH and MAE Congruent Hu Zhi Stupid, BH=AMB BH=AD Yizheng BHA and ADC Congruence (2) with similar triangles cm CE=Mn De=2 3 De=1 2AM Mn=1 3AM AN=2 3AM AM=1 2AC AN=1 3AC

af/fc=an/ac=1/3

The intersection of the two perpendicular bisectors is the position of the tower.

12.Proof that: (1) Because the distance from the points on the perpendicular bisector to the two points is equal, PA=PB

In the same way, there is. pb=pc

So there is pa=pb=pc

2) Because there is Pa=PC, the point P is on the perpendicular bisector of AC.

It can be proved that ABC is similar to ACD, and from this, it can be concluded that DGA is equal to ABC