In RT ABC, ACB 90, CD AB at D, E is the midpoint of AC, and the extension line of ED crosses the CB

Analysis: As long as it can be proved that the triangle PCD is similar to the triangle PDB.

Because cpd= dpb (1).

And because Cd ab is in D, and E is the midpoint of Ac, according to the midline of the hypotenuse of the right triangle is equal to half of the hypotenuse, it can be concluded that EC=ED, then ECD= EDC

And because acp= cdb=90°, then ecd+ acp= edc+ cdb i.e. pcd= pdb(2).

From (1) and (2), it can be concluded that the triangle PCD is similar to the triangle PDB.

So pc pd=pd pb

In a right-angled triangle ACD, because E is the midpoint of the hypotenuse AC, EA=ED, PDB= ADE= CAD= PCD

Thus δpdb δpcb

So pd pb = pc pd, i.e. the square of pd = pbxpc

Because cd=cf, cdf is an isosceles triangle.

So: f= cdf= ade

Because: de ab

So: triangle aed.

a+∠ade=90

Triangle feb.

f+∠b=90

So: a+ ade= f+ b

Again: f= ade

So: a= b

ABC is an isosceles triangle.

Proof: de ab in e

f+∠b=∠eda+∠a=90°

cd=cf∠f=∠cdf

Whereas cdf= eda

f=∠eda

And f+ b= eda+ a=

b= a abc is an isosceles triangle.

Solution: (1) CDF= CF is obtained by CD=CFD, and the extension line of De AB in E, Ed crosses the extension line of BC in F

This gives ade= cdf, so ade= cfd, aed+ ade=90°, cfd= bfd, cfd+ ebf=90°, and ade= cfd

So ead= ebf, i.e. bac= abc, so abc is an isosceles triangle.

2) If cd=cf and f=30°

The extension of de ab to e, ed to the extension of bc to f

CDF= ADE= F=30°, so EAD= EBF=90°-30°=60°, i.e. BAC= ABC=60°

Whereas, BCA=180°-(EAD+EBF)=180°-(60°+60°)=60°

So abc is an equilateral triangle.

Solution: (1).'.'D is the midpoint of AC.

. ad=dc

' af／／ce .'.FAC ace and FDA CDE (to the vertex angle) then the triangle FDA triangle EDC'. af＝ce

2) Quadrilateral AFCE is rectangular.

'AC ef d is the midpoint of AC, then the quadrilaterals with equal diagonals and bisected sides of FD de are rectangulars.

(1) Proof: In ADF and CDE, af be, fad= ecd

and d is the midpoint of ac, ad=cd

adf=∠cde，△adf≌△cde．

af=ce．

2) Solution: If ac=ef, then the quadrilateral afce is a rectangle Proof: From (1) know: af=ce, af ce, the quadrilateral afce is a parallelogram

and ac=ef, the parallelogram afce is a rectangle

1.∵af∥ce

ace=∠caf，∠afe=∠cef

ad=cd

adf≌△cde

af=ce2.AF and CE

AFCE is a parallelogram.

again ac=ef

AFCE is a rectangle (a parallelogram with an equal diagonal is a rectangle.)

(1) Proof:

af∥ce∠afd=∠ced，∠fad=∠ecd

ad=cd△adf≌△cde

af=CE2) quadrilateral AFCE is rectangular.

Proof of ADF CDE

df=deda=dc

The quadrilateral AFCE is a parallelogram.

AC=EF quadrilateral, AFCE is rectangular.